Program4_W

 我现在做的是第四专题编号为1023的试题，具体内容如下所示：

Problem W

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 20   Accepted Submission(s) : 8

Problem Description

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.<br>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).<br>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.<br>

 

Input

There are many test cases:<br>For every case: <br>The first line has two integer N(1<=N<=50,000), M(0<=M<=100,000),separated by a space.<br>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.<br><br>

 

Output

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

 

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

简单题意：

有一个体育馆，座位呈环状，想象下，貌似体育馆都是这样的，每一列有300个座位，按逆时钟方向编号为1~300，假设行数无穷大。

某一天，有N个人（编号为1~N）来到这个体育馆看一场赛事，主办方提出了M个要求，要求的格式是"A B X"，表示的是，假设A坐在编号为i的列，则B必须坐在编号为（i+x）的列上，这些要求里有一些是错误的，只有和前面的要求产生冲突时才算错误，其它都是正确的。程序要输出错误的要求个数。

解题思路：

这是一道比较简单地并查集题目。

（1）弄清题意，找出出现冲突的位置，判断冲突很简单就是当两个人在同一行坐，同时他们到根节点的距离差值正好是他们之间的差值，此时就出现了冲突了。

（2）关键有两个地方，这也是并查集题目的难点，就是压缩集合，和求节点到根的距离。这里压缩集合就很简单了，一个通用的递归。求到跟的距离dist[a]+= dist[tem];dist[rb]=dist[a]+x-dist[b];注意这两行代码，这是核心代码，首先第一行是求出节点a到根的距离。

编写代码：

#include<stdio.h>
int set[50005];
int dist[50005];
int count;

int find(int a)
{
    if(set[a]==a)return a;
    int tem = set[a];
    set[a]=find( set[a]);
    dist[a] += dist[tem];
    return set[a];
}
void merge(int x,int y,int s)
{
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
    {
        set[fy]=fx;
        dist[fy]=dist[x]+s-dist[y];
    }
    else
    {
        if(dist[y]-dist[x]!=s)
        count++;
    }
}
int main()
{
    int n,m,a,b,c;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        count=0;
        for(int i=2;i<=n;i++)
        {
            set[i]=i;
            dist[i]=0;
        }
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            merge(a,b,c);
        }
        printf("%d\n",count);
    }
    return 0;
}