湖南多校对抗5.24
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湖南多校对抗5.24
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據說A,B,C題都比較水這里就不放代碼了
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D:Facility Locations
然而D題是一個腦經急轉彎的題：有m行，n列，每個位置有可能為0，也可能不為0，問最多選K行是不是可以使得每一列都至少有一個0，其中代價c有個約束條件：These costs satisfy a locality property: for two clients j and j’ and two facilities i and i’, we have cij ≤ ci’j + ci’j’ + cij’ .
一看特別像是重復覆蓋的模板，然而稍加分析就會發現由于上面約束的存在，那么任意兩行之間的0的擺放要么完全相同，要么完全錯位，所以最后只要找出任意k個完全錯開的行，判斷是不是每一列都有一個0就可以了= =
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 #define INF 0x3f3f3f3f 16 #define inf (-((LL)1<<40)) 17 #define lson k<<1, L, (L + R)>>1 18 #define rson k<<1|1, ((L + R)>>1) + 1, R 19 #define mem0(a) memset(a,0,sizeof(a)) 20 #define mem1(a) memset(a,-1,sizeof(a)) 21 #define mem(a, b) memset(a, b, sizeof(a)) 22 #define FIN freopen("in.txt", "r", stdin) 23 #define FOUT freopen("out.txt", "w", stdout) 24 #define rep(i, a, b) for(int i = a; i <= b; i ++) 25 #define dec(i, a, b) for(int i = a; i >= b; i --) 26 27 template<class T> T CMP_MIN(T a, T b) { return a < b; } 28 template<class T> T CMP_MAX(T a, T b) { return a > b; } 29 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 30 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 31 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 32 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 33 34 //typedef __int64 LL; 35 typedef long long LL; 36 const int MAXN = 51000; 37 const int MAXM = 110000; 38 const double eps = 1e-4; 39 LL MOD = 1000000007; 40 41 int n, m, k, x; 42 bool vis[110]; 43 44 int main() 45 { 46 while(~scanf("%d %d %d", &n, &m, &k)) { 47 mem0(vis); 48 int time = 0, col = 0; 49 rep (i, 1, n) { 50 int add = 0; 51 rep (j, 1, m) { 52 scanf("%d", &x); 53 if(x || vis[j]) continue; 54 if(!add) time ++; 55 add = 1; 56 vis[j] = 1, col ++; 57 } 58 } 59 puts(col == m && time <= k ? "yes" : "no"); 60 } 61 return 0; 62 } View Code?
E：Repeated Substrings?據說是一個后綴數組的最長公共前綴（LCP）的應用，
賽后又自己寫了一遍后綴數組（其實是把模板打了一遍，模板見鏈接），終于還是把這題A了
題解見鏈接
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 #define INF 0x3f3f3f3f 16 #define inf (-((LL)1<<40)) 17 #define lson k<<1, L, (L + R)>>1 18 #define rson k<<1|1, ((L + R)>>1) + 1, R 19 #define mem0(a) memset(a,0,sizeof(a)) 20 #define mem1(a) memset(a,-1,sizeof(a)) 21 #define mem(a, b) memset(a, b, sizeof(a)) 22 #define FIN freopen("in.txt", "r", stdin) 23 #define FOUT freopen("out.txt", "w", stdout) 24 #define rep(i, a, b) for(int i = a; i <= b; i ++) 25 #define dec(i, a, b) for(int i = a; i >= b; i --) 26 27 template<class T> T CMP_MIN(T a, T b) { return a < b; } 28 template<class T> T CMP_MAX(T a, T b) { return a > b; } 29 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 30 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 31 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 32 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 33 34 //typedef __int64 LL; 35 typedef long long LL; 36 const int MAXN = 110000; 37 const int MAXM = 110000; 38 const double eps = 1e-4; 39 LL MOD = 1000000007; 40 41 struct SufArray { 42 char s[MAXN]; 43 int sa[MAXN], t[MAXN], t2[MAXN], c[MAXN], n, m; 44 int rnk[MAXN], height[MAXN]; 45 int mi[MAXN][20], idxK[MAXN]; 46 47 void init() { 48 mem0(s); 49 mem0(height); 50 } 51 void read_str() { 52 gets(s); 53 m = 128; 54 n = strlen(s); 55 s[n++] = ' '; 56 } 57 void build_sa() { 58 int *x = t, *y = t2; 59 rep (i, 0, m - 1) c[i] = 0; 60 rep (i, 0, n - 1) c[x[i] = s[i]] ++; 61 rep (i, 1, m - 1) c[i] += c[i - 1]; 62 dec (i, n - 1, 0) sa[--c[x[i]]] = i; 63 for(int k = 1; k <= n; k <<= 1) { 64 int p = 0; 65 rep (i, n - k, n - 1) y[p++] = i; 66 rep (i, 0, n - 1) if(sa[i] >= k) y[p++] = sa[i] - k; 67 rep (i, 0, m - 1) c[i] = 0; 68 rep (i, 0, n - 1) c[x[y[i]]] ++; 69 rep (i, 0, m - 1) c[i] += c[i - 1]; 70 dec (i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 71 swap(x, y); 72 p = 1; 73 x[sa[0]] = 0; 74 rep (i, 1, n - 1) { 75 x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k] ? p - 1 : p++; 76 } 77 if(p >= n) break; 78 m = p; 79 } 80 } 81 void get_height() { 82 int k = 0; 83 rep (i, 0, n - 1) rnk[sa[i]] = i; 84 rep (i, 0, n - 1) { 85 if(k) k --; 86 int j = sa[rnk[i] - 1]; 87 while(s[i + k] == s[j + k]) k ++; 88 height[rnk[i]] = k; 89 } 90 } 91 void rmq_init(int *a, int n) { 92 rep (i, 0, n - 1) mi[i][0] = a[i]; 93 for(int j = 1; (1 << j) <= n; j ++) { 94 for(int i = 0; i + (1<<j) - 1 < n; i ++) { 95 mi[i][j] = min(mi[i][j - 1], mi[i + (1 << (j - 1))][j - 1]); 96 } 97 } 98 rep (len, 1, n) { 99 idxK[len] = 0; 100 while((1 << (idxK[len] + 1)) <= len) idxK[len] ++; 101 } 102 } 103 int rmq_min(int l, int r) { 104 int len = r - l + 1, k = idxK[len]; 105 return min(mi[l][k], mi[r - (1 << k) + 1][k]); 106 } 107 void lcp_init() { 108 get_height(); 109 rmq_init(height, n); 110 } 111 int get_lcp(int a, int b) { 112 if(a == b) return n - a - 1; 113 return rmq_min(min(rnk[a], rnk[b]) + 1, max(rnk[a], rnk[b])); 114 } 115 void solve() { 116 get_height(); 117 LL ans = 0, pre = 0; 118 rep (i, 1, n - 1) { 119 if(height[i] > pre) ans += height[i] - pre; 120 pre = height[i]; 121 } 122 cout << ans << endl; 123 } 124 }; 125 126 int T; 127 SufArray sa; 128 129 int main() 130 { 131 while(~scanf("%d%*c", &T)) while(T--){ 132 sa.init(); 133 sa.read_str(); 134 sa.build_sa(); 135 sa.solve(); 136 } 137 return 0; 138 } 139 /************************************************************** 140 Problem: 1632 141 User: csust_Rush 142 Language: C++ 143 Result: Accepted 144 Time:880 ms 145 Memory:13192 kb 146 ****************************************************************/ View Code?
F：Landline Telephone Network題意就是求一顆最小生成樹，但是任意兩個點之間的唯一路徑上不能通過任何一個 壞點 ?，題目告訴你哪些是壞點
其實就是先不管壞點，求一遍最小生成樹，然后把壞點加進去（注意細節地方）
1 #include <stdio.h> 2 #include <iostream> 3 #include <cmath> 4 #include <cstdlib> 5 #include <vector> 6 #include <algorithm> 7 #include <cstring> 8 #define mem0(a) memset(a, 0, sizeof(a)) 9 #define rep(i,a, b) for(int i = a; i <= b; i ++) 10 using namespace std; 11 12 int fa[1100]; 13 struct Edge { 14 int u, v, w; 15 Edge(int _u = 0, int _v = 0, int _w = 0) { 16 u = _u; v = _v; w = _w; 17 } 18 bool operator < (const Edge &A) const { 19 return w < A.w; 20 } 21 }e[110000]; 22 int n, m, p, x, b[1100], u, v, w; 23 int t[1100]; 24 25 int findFa(int x) { 26 return x == fa[x] ? x : fa[x] = findFa(fa[x]); 27 } 28 29 int main() 30 { 31 //freopen("in.txt", "r", stdin); 32 while(~scanf("%d %d %d", &n, &m, &p) && n) { 33 mem0(b); mem0(t); 34 rep (i, 1, n) fa[i] = i; 35 rep (i, 1, p) scanf("%d", &x), b[x] = 1; 36 rep (i, 1, m) { 37 scanf("%d %d %d", &u, &v, &w); 38 e[i] = Edge(u, v, w); 39 } 40 if(n == 2) { 41 cout << w << endl; 42 continue; 43 } 44 sort(e + 1, e + m + 1); 45 int ans = 0; 46 rep (i, 1, m) { 47 u = e[i].u, v = e[i].v, w = e[i].w; 48 if(b[u] || b[v]) continue; 49 int x = findFa(u), y = findFa(v); 50 if(x != y) { 51 ans += w; 52 fa[x] = y; 53 } 54 } 55 int imposs = 0; 56 rep (i, 1, m) { 57 u = e[i].u, v = e[i].v, w = e[i].w; 58 if(!(b[u] ^ b[v])) continue; 59 t[u]++; t[v] ++; 60 if(b[u] && t[u] >= 2) continue; 61 if(b[v] && t[v] >= 2) continue; 62 int x = findFa(u), y = findFa(v); 63 if(x != y) { 64 ans += w; 65 fa[x] = y; 66 } 67 } 68 int cnt = 0; 69 rep (i, 1, n) if(findFa(fa[i]) == i) {cnt ++; } 70 if(imposs || cnt > 1) puts("impossible"); 71 else cout << ans << endl; 72 } 73 return 0; 74 } 75 76 /************************************************************** 77 Problem: 1633 78 User: csust_Rush 79 Language: C++ 80 Result: Accepted 81 Time:84 ms 82 Memory:2792 kb 83 ****************************************************************/ View Code?
GAquarium Tank簡單計算幾何
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 #define INF 0x3f3f3f3f 16 #define inf (-((LL)1<<40)) 17 #define lson k<<1, L, (L + R)>>1 18 #define rson k<<1|1, ((L + R)>>1) + 1, R 19 #define mem0(a) memset(a,0,sizeof(a)) 20 #define mem1(a) memset(a,-1,sizeof(a)) 21 #define mem(a, b) memset(a, b, sizeof(a)) 22 #define FIN freopen("in.txt", "r", stdin) 23 #define FOUT freopen("out.txt", "w", stdout) 24 #define rep(i, a, b) for(int i = a; i <= b; i ++) 25 #define dec(i, a, b) for(int i = a; i >= b; i --) 26 27 template<class T> T CMP_MIN(T a, T b) { return a < b; } 28 template<class T> T CMP_MAX(T a, T b) { return a > b; } 29 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 30 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 31 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 32 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 33 34 //typedef __int64 LL; 35 typedef long long LL; 36 const int MAXN = 110000; 37 const int MAXM = 110000; 38 const double eps = 1e-6; 39 LL MOD = 1000000007; 40 41 int n; 42 double D, L; 43 struct Point { 44 double x, y; 45 Point(double _x = 0, double _y = 0) { 46 x = _x; y = _y; 47 } 48 }; 49 50 double cross(Point a, Point b) { 51 return a.x * b.y - a.y * b.x; 52 } 53 54 struct Polygon { 55 Point p[MAXN]; 56 int cnt; 57 double get_area() { 58 int pre = cnt; 59 double ans = 0; 60 rep (i, 1, cnt) { 61 ans += cross(p[pre], p[i]); 62 pre = i; 63 } 64 return fabs(ans / 2); 65 } 66 67 }pol, po; 68 69 Point get_point(Point a, Point b, double y0) { 70 if(fabs(a.x - b.x) < eps) return Point(a.x, y0); 71 double bi = (y0 - a.y) / (b.y - a.y); 72 return Point(a.x + bi * (b.x - a.x), a.y + bi * (b.y - a.y)); 73 } 74 75 double find_area(double y0) { 76 int cnt = 0, pre = po.cnt; 77 rep (i, 1, po.cnt) { 78 if((y0 - po.p[pre].y) * (y0 - po.p[i].y) <= 0) { 79 pol.p[++cnt] = get_point(po.p[pre], po.p[i], y0); 80 } 81 if(po.p[i].y < y0) pol.p[++cnt] = po.p[i]; 82 pre = i; 83 } 84 pol.cnt = cnt; 85 return pol.get_area(); 86 } 87 88 int main() 89 { 90 //FIN; 91 while(~scanf("%d", &po.cnt)) { 92 cin >> D >> L; 93 L *= 1000; 94 double mx_y = 0.0; 95 rep (i, 1, po.cnt) { 96 scanf("%lf %lf", &po.p[i].x, &po.p[i].y); 97 mx_y = max(mx_y, po.p[i].y); 98 } 99 double low = 0.0, high = mx_y; 100 while(high - low > eps) { 101 double mid = (low + high) / 2.0; 102 if(find_area(mid) * D < L) low = mid; 103 else high = mid; 104 } 105 printf("%.2lf\n", low); 106 } 107 return 0; 108 } 109 110 /************************************************************** 111 Problem: 1634 112 User: csust_Rush 113 Language: C++ 114 Result: Accepted 115 Time:12 ms 116 Memory:4920 kb 117 ****************************************************************/ View Code?
H：Restaurant Ratings
一個數位的簡單DP，DP[i][j]表示i個人打j分的方案數
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 #define INF 0x3f3f3f3f 16 #define inf (-((LL)1<<40)) 17 #define lson k<<1, L, (L + R)>>1 18 #define rson k<<1|1, ((L + R)>>1) + 1, R 19 #define mem0(a) memset(a,0,sizeof(a)) 20 #define mem1(a) memset(a,-1,sizeof(a)) 21 #define mem(a, b) memset(a, b, sizeof(a)) 22 #define FIN freopen("in.txt", "r", stdin) 23 #define FOUT freopen("out.txt", "w", stdout) 24 #define rep(i, a, b) for(int i = a; i <= b; i ++) 25 #define dec(i, a, b) for(int i = a; i >= b; i --) 26 27 template<class T> T CMP_MIN(T a, T b) { return a < b; } 28 template<class T> T CMP_MAX(T a, T b) { return a > b; } 29 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 30 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 31 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 32 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 33 34 //typedef __int64 LL; 35 typedef long long LL; 36 const int MAXN = 51000; 37 const int MAXM = 110000; 38 const double eps = 1e-4; 39 LL MOD = 1000000007; 40 41 LL dp[110][110]; 42 43 void init_dp(int n) { 44 dp[0][0] = 1; 45 rep (i, 1, n) { 46 rep (j, 0, n) { 47 rep (k, 0, j) { 48 dp[i][j] += dp[i - 1][j - k]; 49 } 50 } 51 } 52 } 53 54 int n, a[110]; 55 56 int main() 57 { 58 init_dp(100); 59 while(~scanf("%d", &n) && n) { 60 int sum = 0; 61 rep (i, 1, n) { 62 scanf("%d", &a[i]); 63 sum += a[i]; 64 } 65 LL ans = 1; 66 rep (i, 0, sum - 1) { 67 ans += dp[n][i]; 68 } 69 rep (i, 1, n) { 70 rep (j, 0, a[i] - 1) { 71 ans += dp[n - i][sum - j]; 72 } 73 sum -= a[i]; 74 } 75 cout << ans << endl; 76 } 77 return 0; 78 } 79 80 /************************************************************** 81 Problem: 1635 82 User: csust_Rush 83 Language: C++ 84 Result: Accepted 85 Time:16 ms 86 Memory:1580 kb 87 ****************************************************************/ View Code?
I：Locked Treasure
據說是看樣例找規律，然后發現ans=C(n, m-1)，不過也有類似的完美證明
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J：Yet Satisfiability Again!
比賽時最后幾分鐘用暴力交了一發，發現AC了，賽后看其他人時間，大概都是暴利做的
最后干脆連題解都是用暴力啊= =
1 #include <stdio.h> 2 #include <iostream> 3 #include <cmath> 4 #include <cstdlib> 5 #include <vector> 6 #include <algorithm> 7 #include <cstring> 8 #define mem0(a) memset(a, 0, sizeof(a)) 9 #define rep(i,a, b) for(int i = a; i <= b; i ++) 10 using namespace std; 11 12 int t, n, m; 13 struct Node { 14 int a[21], f[21], cnt; 15 bool judge(int s) { 16 int ans = 0, p = 0; 17 while(p < cnt && !ans) { 18 ans |= a[p] ^ (((1<<f[p]) & s) ? 1 : 0); 19 p ++; 20 } 21 return ans; 22 } 23 }mt[110]; 24 char s[111100]; 25 26 void handle(char *s, int id) { 27 int len = strlen(s), cnt = 0; 28 rep (i, 0, len - 1) { 29 if(s[i] == 'X') { 30 int num = 0, p = i; 31 while(isdigit(s[++p])) num = num * 10 + s[p] - '0'; 32 mt[id].a[cnt] = i > 0 && s[i - 1] == '~'; 33 mt[id].f[cnt++] = num - 1; 34 } 35 } 36 mt[id].cnt = cnt; 37 } 38 39 int main() 40 { 41 //freopen("in.txt", "r", stdin); 42 while(cin >> t) while(t--) { 43 scanf("%d %d", &n, &m); 44 gets(s); 45 rep (i, 1, m) { 46 gets(s); 47 handle(s, i); 48 } 49 int ok = 0, h = (1 << n) - 1; 50 rep (s, 0, h) { 51 int p = 1; 52 rep (i, 1, m) { 53 p &= mt[i].judge(s); 54 if(!p) break; 55 } 56 ok |= p; 57 if(ok) { 58 break; 59 } 60 } 61 if(ok) puts("satisfiable"); 62 else puts("unsatisfiable"); 63 } 64 return 0; 65 } 66 67 /************************************************************** 68 Problem: 1637 69 User: csust_Rush 70 Language: C++ 71 Result: Accepted 72 Time:1768 ms 73 Memory:1612 kb 74 ****************************************************************/ View Code?
K：Continued Fraction
做了兩次了，第一次以為超LL，后來隊友用BigInt（自己寫的模板）過了
后來又聽說LL可以過，再回頭看自己代碼，發現有用LL*LL，這必然超啊。。。改了后發現立馬AC
1 #include <stdio.h> 2 #include <iostream> 3 #include <cmath> 4 #include <cstdlib> 5 using namespace std; 6 7 typedef long long LL; 8 9 int n, m; 10 LL a1[100], a2[100]; 11 12 LL gcd(LL a, LL b) { 13 return b == 0 ? a : gcd(b, a % b); 14 } 15 16 void dfs1(LL &a, LL &b, LL *arr, int i, int n) 17 { 18 if(i >= n - 1) { 19 a = 1; b = arr[i]; return ; 20 } 21 dfs1(a, b, arr, i + 1, n); 22 LL tmp = a; 23 a = b; b = arr[i] * b + tmp; 24 LL g = gcd(a, b); 25 a /= g; b /= g; 26 } 27 28 void add(LL A1, LL B1, LL A2, LL B2, LL &x, LL &y) { 29 x = A1 * B2 + A2 * B1; 30 y = B1 * B2; 31 LL g = gcd(x, y); 32 x /= g; y /= g; 33 } 34 void sub(LL A1, LL B1, LL A2, LL B2, LL &x, LL &y) { 35 x = A1 * B2 - A2 * B1; 36 y = B1 * B2; 37 LL g = gcd(x, y); 38 x /= g; y /= g; 39 } 40 void mul(LL A1, LL B1, LL A2, LL B2, LL &x, LL &y) { 41 x = A1 * A2; y = B1 * B2; 42 LL g = gcd(x, y); 43 x /= g; y /= g; 44 } 45 void div(LL A1, LL B1, LL A2, LL B2, LL &x, LL &y) { 46 x = A1 * B2; y = A2 * B1; 47 LL g = gcd(x, y); 48 x /= g; y /= g; 49 } 50 51 void print(LL x, LL y) { 52 if (x > 0 && y < 0 || x < 0 && y > 0) { 53 if (x % y != 0) cout<< x / y - 1 << " "; 54 else { 55 cout << x / y << endl;; 56 return; 57 } 58 x = abs(y) - abs(x) % abs(y); 59 print(abs(y), x); 60 return ; 61 } 62 cout << x / y; 63 if(x % y) { 64 printf(" "); 65 print(y, x % y); 66 } 67 else printf("\n"); 68 } 69 70 int main() 71 { 72 //freopen("in.txt", "r", stdin); 73 int cas = 0; 74 while(~scanf("%d %d", &n, &m)) { 75 for(int i = 0; i < n; i ++) cin >> a1[i]; 76 for(int i = 0; i < m; i ++) cin >> a2[i]; 77 LL A1 = 0, B1 = 1, A2 = 0, B2 = 1; 78 if (n > 1) dfs1(A1, B1, a1, 1, n); 79 if (m > 1) dfs1(A2, B2, a2, 1, m); 80 A1 = a1[0] * B1 + A1; 81 A2 = a2[0] * B2 + A2; 82 LL x, y; 83 //printf("Case %d:\n", ++cas); 84 add(A1, B1, A2, B2, x, y); print(x, y); 85 sub(A1, B1, A2, B2, x, y); print(x, y); 86 mul(A1, B1, A2, B2, x, y); print(x, y); 87 div(A1, B1, A2, B2, x, y); print(x, y); 88 } 89 return 0; 90 } 91 View Code?
轉載于:https://www.cnblogs.com/gj-Acit/p/4527661.html
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