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The Experience of Love

Time Limit: 4000/2000 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 221????Accepted Submission(s): 91

Problem Description A girl named Gorwin and a boy named Vivin is a couple. They arrived at a country named LOVE. The country consisting of N cities and only N?1 edges (just like a tree), every edge has a value means the distance of two cities. They select two cities to live，Gorwin living in a city and Vivin living in another. First date, Gorwin go to visit Vivin, she would write down the longest edge on this path(maxValue).Second date, Vivin go to Gorwin, he would write down the shortest edge on this path(minValue),then calculate the result of maxValue subtracts minValue as the experience of love, and then reselect two cities to live and calculate new experience of love, repeat again and again.

Please help them to calculate the sum of all experience of love after they have selected all cases.

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Input There will be about 5 cases in the input file.
For each test case the first line is a integer N , Then follows n?1 lines, each line contains three integers a , b , and c , indicating there is a edge connects city a and city b with distance c .

[Technical Specification]
1<N<=150000,1<=a,b<=n,1<=c<=109

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Output For each case，the output should occupies exactly one line. The output format is Case #x: answer, here x is the data number, answer is the sum of experience of love.

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Sample Input 3 1 2 1 2 3 2 5 1 2 2 2 3 5 2 4 7 3 5 4

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Sample Output Case #1: 1 Case #2: 17 Hint huge input,fast IO method is recommended. In the first sample: The maxValue is 1 and minValue is 1 when they select city 1 and city 2, the experience of love is 0. The maxValue is 2 and minValue is 2 when they select city 2 and city 3, the experience of love is 0. The maxValue is 2 and minValue is 1 when they select city 1 and city 3, the experience of love is 1. so the sum of all experience is 1;

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Source Valentine's Day Round

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Recommend hujie???|???We have carefully selected several similar problems for you:??5177?5175?5173?5170?5169?

轉(zhuǎn)一發(fā)官方題解：http://bestcoder.hdu.edu.cn/

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題意：給一棵樹，求任意｛兩點路徑上的最大邊權(quán)值-最小邊權(quán)值｝的總和。 解法：sigma(maxVal[i]?minVal[i])=sigma(maxVal)?sigma(minVal) ;所以我們分別求所有兩點路徑上的最大值的和，還有最小值的和。再相減就可以了。求最大值的和的方法用帶權(quán)并查集，把邊按權(quán)值從小到大排序，一條邊一條邊的算，當我們算第i 條邊的時候權(quán)值為wi ,兩點是ui,vi ,前面加入的邊權(quán)值一定是小于等于當前wi 的，假設(shè)與ui 連通的點有a 個，與vi 連通的點有b 個，那么在a 個中選一個，在b 個中選一個，這兩個點的路徑上最大值一定是wi ,一共有a?b 個選法，愛情經(jīng)驗值為a?b?wi 。求最小值的和的方法類似。 槽點： 一：這題做數(shù)據(jù)的時候突然想到的把數(shù)據(jù)范圍設(shè)在 unsigned long long 范圍內(nèi)，要爆 long long，這樣選手在wa了之后可能心態(tài)不好找不到這個槽點，當是鍛煉大家的心態(tài)和出現(xiàn)wa時的找錯能力了，把這放在pretest..很良心的。 二，并查集的時候，用是遞歸的需要擴棧，一般上10w 的遞歸都需要，所以看見有幾個FST在棧溢出的，好桑心。

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12957565

2015-02-16 11:18:47

Accepted

5176

842MS

6820K

2033 B

G++

czy

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1 #include<iostream> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cstdio> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<string> 12 13 #define N 150005 14 #define M 10005 15 //#define mod 10000007 16 //#define p 10000007 17 #define mod2 1000000000 18 #define ll long long 19 #define ull unsigned long long 20 #define LL long long 21 #define eps 1e-6 22 //#define inf 2147483647 23 #define maxi(a,b) (a)>(b)? (a) : (b) 24 #define mini(a,b) (a)<(b)? (a) : (b) 25 26 using namespace std; 27 28 int n; 29 int f[N]; 30 ull cou[N]; 31 ull suma,sumi; 32 int cnt; 33 34 typedef struct 35 { 36 int a; 37 int b; 38 ull c; 39 }PP; 40 41 PP p[N]; 42 43 bool cmp(PP x,PP y) 44 { 45 return x.c<y.c; 46 } 47 48 int find(int x) 49 { 50 int fa; 51 if(x!=f[x]) 52 { 53 fa=find(f[x]); 54 f[x]=fa; 55 } 56 return f[x]; 57 } 58 59 void merge(int x,int y) 60 { 61 int a,b; 62 a=find(x); 63 b=find(y); 64 if(a==b) return; 65 f[b]=a; 66 cou[a]=cou[a]+cou[b]; 67 } 68 69 void ini() 70 { 71 suma=sumi=0; 72 int i; 73 for(i=0;i<=n;i++){ 74 f[i]=i; 75 cou[i]=1; 76 } 77 for(i=1;i<n;i++){ 78 scanf("%d%d%I64u",&p[i].a,&p[i].b,&p[i].c); 79 } 80 sort(p+1,p+n,cmp); 81 } 82 83 void solve() 84 { 85 int i; 86 int aa,bb; 87 for(i=1;i<n;i++){ 88 aa=find(p[i].a); 89 bb=find(p[i].b); 90 suma+=cou[aa]*cou[bb]*p[i].c; 91 merge(p[i].a,p[i].b); 92 } 93 for(i=0;i<=n;i++){ 94 f[i]=i; 95 cou[i]=1; 96 } 97 for(i=n-1;i>=1;i--){ 98 aa=find(p[i].a); 99 bb=find(p[i].b); 100 sumi+=cou[aa]*cou[bb]*p[i].c; 101 merge(p[i].a,p[i].b); 102 } 103 } 104 105 void out() 106 { 107 printf("Case #%d: %I64u\n",cnt,suma-sumi); 108 cnt++; 109 } 110 111 int main() 112 { 113 cnt=1; 114 //freopen("data.in","r",stdin); 115 //freopen("data.out","w",stdout); 116 //scanf("%d",&T); 117 //for(int ccnt=1;ccnt<=T;ccnt++) 118 //while(T--) 119 //scanf("%d%d",&n,&m); 120 while(scanf("%d",&n)!=EOF) 121 { 122 ini(); 123 solve(); 124 out(); 125 } 126 return 0; 127 }

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轉(zhuǎn)載于:https://www.cnblogs.com/njczy2010/p/4293905.html

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