2016-2017 ACM-ICPC CHINA-Final

A. Number Theory Problem（規律、簽到）

#include <bits/stdc++.h>using namespace std;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);int T, n, cas = 0;scanf("%d", &T);while (T--) {scanf("%d", &n);printf("Case #%d: %d\n", ++cas, n / 3);}return 0; }

C. Mr. Panda and Strips

給定一個長度為$n,(1\le n\le 1000)$的數組，里面元素為$c_i,(1\le c_i\le 10^5)$，

問，在里面選出兩端不相交的子串（可為空），拼接使得拼接后的子串里面沒有重復元素出現，輸出這個子串的最大長度。

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10;int pre[N], suc[N], pos[N], a[N], n;set<int> st1, st2;int maxllen(int p, int l, int llen) {auto it = st2.end();while ((it--) != st2.begin()) {if (pos[a[*it]] < p) {return l - *it;}}return llen; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);int T, cas = 0;scanf("%d", &T);while (T--) {printf("Case #%d: ", ++cas);scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);}for (int i = 1; i < N; i++) {pos[i] = 0;}for (int i = 1; i <= n; i++) {pre[i] = pos[a[i]];pos[a[i]] = i;}for (int i = 1; i < N; i++) {pos[i] = n + 1;}for (int i = n; i >= 1; i--) {suc[i] = pos[a[i]];pos[a[i]] = i;}int ans = 1;st1.clear();for (int i = n, rlen = 1; i >= 1; i--, rlen++) {while (suc[i] <= i + rlen - 1) {st1.erase(a[i + rlen - 1]);rlen--;}st1.insert(a[i]);pos[a[i]] = i;st2.clear();for (int j = 1, llen = 1; j < i; j++, llen++) {while (pre[j] >= j - llen + 1) {st2.erase(j - llen + 1);llen--;}if (st1.count(a[j])) {st2.insert(j);}if (!st2.size()) {ans = max(ans, llen + rlen);}else {ans = max(ans, j - *(--st2.end()) + rlen);}for (auto it : st2) {ans = max(ans, maxllen(pos[a[it]], j, llen) + pos[a[it]] - i);}}}printf("%d\n", ans);}return 0; }

D. Ice Cream Tower（二分、貪心）

二分答案，然后貪心 judge 即可。

#include <bits/stdc++.h>using namespace std;const int N = 3e5 + 10;int n, k;long long a[N], b[N];bool judge(int x) {int num = 1, tot = 1;for (int i = 1; i <= x; i++) {b[i] = a[i];}for (int i = x + 1; i <= n; i++) {if (2 * a[i] <= b[tot]) {b[tot++] = a[i];}if (tot == x + 1) {tot = 1;num++;}}return num >= k; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);int T, cas = 0;scanf("%d", &T);while (T--) {scanf("%d %d", &n, &k);for (int i = 1; i <= n; i++) {scanf("%lld", &a[i]);}sort(a + 1, a + 1 + n, greater<long long> ());int l = 0, r = n;while (l < r) {int mid = l + r + 1 >> 1;if (judge(mid)) {l = mid;}else {r = mid - 1;}}printf("Case #%d: %d\n", ++cas, l);}return 0; }

E. Bet（思維、貪心）

設我們總共投入$S$元，則對于買入的比賽，我們必須投$x_i$元，有$x_i(1+\frac{B_i}{A_i})=S$，$x_i=\frac{A_{i}S}{A_i+B_i}$，要使$\sum _i\frac{A_{i}S}{A_i+B_i}<S$

則$\sum _i\frac{A_i}{A_i+B_i}<1$，考慮對$\frac{A_i}{A_i+B_i}$排一個序，貪心求解即可。

#include <bits/stdc++.h>using namespace std;struct Res {long double a, b;void read() {scanf("%Lf:%Lf", &a, &b);}bool operator < (const Res &t) const {return a * (t.a + t.b) < t.a * (a + b);} }a[110];int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);int T, n, cas = 0;scanf("%d", &T);while (T--) {scanf("%d", &n);for (int i = 1; i <= n; i++) {a[i].read();}sort(a + 1, a + 1 + n);int ans = 0;long double cur = 0;for (int i = 1; i <= n; i++) {if (cur + a[i].a / (a[i].a + a[i].b) < 1.0) {cur += a[i].a / (a[i].a + a[i].b);ans++;}}printf("Case #%d: %d\n", ++cas, ans);}return 0; }

G. Pandaria （Kruskal 重構樹、線段樹合并）

給定一個有$n$條邊的無向連通圖，每條邊有對應的邊權，每個點有一個顏色，

問從一個點出發，經過不超過$w$的邊權，所能到達的點中，顏色出現次數做多且顏色編號最小的是什么顏色。

不超過某個權值所能到達的點，由此我們可以考慮建立升序$kruskal$重構樹，然后從某個點倍增往上跳，直到不能跳為止，

這個時候我們所在的點的子樹所包含的點就是我們能夠到達的點了，

考慮用權值線段樹來維護每一個點所代表的子樹的信息，更新的時候我們只要往上進行線段樹合并就行了。

#include <bits/stdc++.h>using namespace std;const int N = 2e5 + 10;int n, nn, m, Q, ff[N], value[N], color[N], fa[N][21], ans[N];int root[N], ls[N << 5], rs[N << 5], maxn[N << 5], num;vector<int> G[N];struct Res {int u, v, w;void read() {scanf("%d %d %d", &u, &v, &w);}bool operator < (const Res &t) const {return w < t.w;} }edge[N];void push_up(int rt) {maxn[rt] = max(maxn[ls[rt]], maxn[rs[rt]]); }void update(int &rt, int l, int r, int x, int v) {if (!rt) {rt = ++num;}if (l == r) {maxn[rt] += v;return ;}int mid = l + r >> 1;if (x <= mid) {update(ls[rt], l, mid, x, v);}else {update(rs[rt], mid + 1, r, x, v);}push_up(rt); }int merge(int x, int y, int l, int r) {if (x == 0 || y == 0) {return x | y;}if (l == r) {maxn[x] += maxn[y];return x;}int mid = l + r >> 1;ls[x] = merge(ls[x], ls[y], l, mid);rs[x] = merge(rs[x], rs[y], mid + 1, r);push_up(x);return x; }int query(int rt, int l, int r) {if (l == r) {return l;}int mid = l + r >> 1;if (maxn[ls[rt]] == maxn[rt]) {return query(ls[rt], l, mid);}else {return query(rs[rt], mid + 1, r);} }void dfs(int rt, int f) {fa[rt][0] = f;for (int i = 1; i <= 20; i++) {fa[rt][i] = fa[fa[rt][i - 1]][i - 1];}if (rt <= n) {update(root[rt], 1, n, color[rt], 1);}for (int to : G[rt]) {if (to == f) {continue;}dfs(to, rt);root[rt] = merge(root[rt], root[to], 1, n);}ans[rt] = query(root[rt], 1, n); }int find(int rt) {return ff[rt] == rt ? rt : ff[rt] = find(ff[rt]); }void kruskal() {sort(edge + 1, edge + 1 + m);for (int i = 1; i < 2 * n; i++) {ff[i] = i;}for (int i = 1, cur = 1; i <= m && cur < n; i++) {int u = find(edge[i].u), v = find(edge[i].v);if (u ^ v) {nn++, cur++;ff[u] = ff[v] = nn;G[nn].push_back(u), G[nn].push_back(v);value[nn] = edge[i].w;if (u <= n) {value[u] = edge[i].w;}if (v <= n) {value[v] = edge[i].v;}}}for (int i = 1; i <= num; i++) {ls[i] = rs[i] = maxn[i] = 0;}for (int i = 1; i <= nn; i++) {root[i] = 0;}num = 0;dfs(nn, 0);for (int i = 1; i <= nn; i++) {G[i].clear();} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T;scanf("%d", &T);for (int cas = 1; cas <= T; cas++) {printf("Case #%d:\n", cas);scanf("%d %d", &n, &m);nn = n;for (int i = 1; i <= n; i++) {scanf("%d", &color[i]);}for (int i = 1; i <= m; i++) {edge[i].read();}kruskal();scanf("%d", &Q);int res = 0;while (Q--) {int u, x;scanf("%d %d", &u, &x);u ^= res, x ^= res;for (int i = 20; i >= 0; i--) {if (fa[u][i] && value[fa[u][i]] <= x) {u = fa[u][i];}}res = ans[u];printf("%d\n", res);}}return 0; }

H. Great Cells

給定一個$n\times m$的網格，往里面填數，數字范圍為$[1,k]$，定義 great cell 是其所在行列的嚴格最大值，

$A_g$，表示如果有$g$個 great cell 的方案，要我們求$\sum _{g=0}^{n\times m}(g+1)A_g$。
$$\sum _{g=0}^{n\times m}{A}_g+\sum _{g=0}^{n\times m}g{A}_g$$
容易發現前項答案為$k^{n\times m}$，考慮后項如何求解，

L. World Cup（枚舉、簽到）

三進制枚舉一下每一場比賽的勝負，然后用 map 存一下，然后判斷答案即可。

#include <bits/stdc++.h>using namespace std;map<vector<int>, int> mp;const int a[6] = {1, 1, 1, 2, 2, 3}, b[6] = {2, 3, 4, 3, 4, 4};void init() {for (int i = 0; i < 729; i++) {vector<int> vt(5, 0);int cur = i;for (int j = 0; j < 6; j++) {if (cur % 3 == 0) {vt[a[j]]++, vt[b[j]]++;}else if (cur % 3 == 1) {vt[a[j]] += 3;}else {vt[b[j]] += 3;}cur /= 3;}mp[vt]++;} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);init();int T, cas = 0;scanf("%d", &T);while (T--) {vector<int> vt(5, 0);for (int i = 1; i <= 4; i++) {scanf("%d", &vt[i]);}printf("Case #%d: ", ++cas);if (!mp.count(vt)) {puts("Wrong Scoreboard");}else if (mp[vt] == 1) {puts("Yes");}else {puts("No");}}return 0; }

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