題意：

觀眾席圍成一圈。列的總數是300，編號為1–300，順時針計數，我們假設行的數量是無限的。將有N個人去那里。他對這些座位提出了要求：這意味著編號A的順時針X距離坐著編號B。例如：A在第4列，X是2，那么B必須在第6列（6=4+2）?，F在你的任務是判斷請求是否正確。

題目：

In 12th Zhejiang College Students Games 2007, there was a new stadium built in Zhejiang Normal University. It was a modern stadium which could hold thousands of people. The audience Seats made a circle. The total number of columns were 300 numbered 1–300, counted clockwise, we assume the number of rows were infinite.
These days, Busoniya want to hold a large-scale theatrical performance in this stadium. There will be N people go there numbered 1–N. Busoniya has Reserved several seats. To make it funny, he makes M requests for these seats: A B X, which means people numbered B must seat clockwise X distance from people numbered A. For example: A is in column 4th and X is 2, then B must in column 6th (6=4+2).
Now your task is to judge weather the request is correct or not. The rule of your judgement is easy: when a new request has conflicts against the foregoing ones then we define it as incorrect, otherwise it is correct. Please find out all the incorrect requests and count them as R.

Input

There are many test cases:
For every case:
The first line has two integer N(1<=N<=50,000), M(0<=M<=100,000),separated by a space.
Then M lines follow, each line has 3 integer A(1<=A<=N), B(1<=B<=N), X(0<=X<300) (A!=B), separated by a space.

Output

For every case:
Output R, represents the number of incorrect request.

Sample Input

10 10
1 2 150
3 4 200
1 5 270
2 6 200
6 5 80
4 7 150
8 9 100
4 8 50
1 7 100
9 2 100

Sample Output

2

Hint

Hint:
（PS： the 5th and 10th requests are incorrect）

思路

1.由于題中明確給出，編號A的順時針X距離坐著編號B，故x<300，當我們用并查集找到一個祖宗節點時，我們可以將該點看作原點，將圈拉成直線看待。即可忽略成環。

2.b->a, a離”原點更近“

int f(int x) {if(x!=dp[x]){int t=dp[x];/*因為在遞歸找祖先的過程中，dp[x]的值會改變，而我們我們只需要找到距離x最近的點，即可得到dp[x]到祖宗節點的距離。所以要記錄t的值*/dp[x]=f(dp[x]);num[x]+=num[t];}return dp[x]; }

dp[b]->dp[a]

dp[v]=u;num[v]=num[a]+x-num[b];

AC代碼

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int M=5e4+10; int dp[M],num[M]; int f(int x) {if(x!=dp[x]){int t=dp[x];dp[x]=f(dp[x]);/*因為在遞歸找祖先的過程中，dp[x]的值會改變，而我們我們只需要找到距離x最近的點，即可得到dp[x]到祖宗節點的距離。所以要記錄t的值*/num[x]+=num[t];}return dp[x]; } bool dfs(int a,int b,int x) {int u=f(a),v=f(b);/*a->u,b->v*/if(u==v){if(num[a]+x!=num[b])/*b->a,又起點為父節點*/return true;return false;}dp[v]=u;num[v]=num[a]+x-num[b];return false; } int main() {int m,n;while(~scanf("%d%d",&m,&n)){int ans=0;for(int i=0;i<m;i++){dp[i]=i;num[i]=0;}while(n--){int a,b,x;/*設起點為祖宗節點，則都指向起點b->a*/scanf("%d%d%d",&a,&b,&x);if(dfs(a,b,x))ans++;}printf("%d\n",ans);}return 0; }

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