并查集：people on a line

People on a Line

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Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

There are N people standing on the x-axis. Let the coordinate of Person i be xi. For every i, xi is an integer between 0 and 109 (inclusive). It is possible that more than one person is standing at the same coordinate.

You will given M pieces of information regarding the positions of these people. The i-th piece of information has the form (Li,Ri,Di). This means that Person Ri is to the right of Person Li by Di units of distance, that is, xRi−xLi=Di holds.

It turns out that some of these M pieces of information may be incorrect. Determine if there exists a set of values (x1,x2,…,xN) that is consistent with the given pieces of information.

Constraints

• 1≤N≤100 000
• 0≤M≤200 000
• 1≤Li,Ri≤N (1≤i≤M)
• 0≤Di≤10 000 (1≤i≤M)
• Li≠Ri (1≤i≤M)
• If i≠j, then (Li,Ri)≠(Lj,Rj) and (Li,Ri)≠(Rj,Lj).
• Di are integers.

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Input

Input is given from Standard Input in the following format:

N M
L1 R1 D1
L2 R2 D2
:
LM RM DM

Output

If there exists a set of values (x1,x2,…,xN) that is consistent with all given pieces of information, print Yes; if it does not exist, print No.

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Sample Input 1

Copy

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Copy

Yes

Some possible sets of values (x1,x2,x3) are (0,1,2) and (101,102,103).

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Sample Input 2

Copy

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

Copy

No

If the first two pieces of information are correct, x3−x1=2 holds, which is contradictory to the last piece of information.

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Sample Input 3

Copy

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Copy

Yes

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Sample Input 4

Copy

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

Copy

No

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Sample Input 5

Copy

100 0

Sample Output 5

Copy

Yes

题目的意思是：给你多组数据每组三个数，分别是哪个点和哪个点之间的距离，问你是否有这样一组数满足所有的组数；

题目用并查集做，每加一组数据就进行一次查询和合并，二个数组，一个存每个点的祖先是谁，还有一个存的是这个点到祖先的距离；

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include<cstring>
#define max(x,y) x>y?x:y
using namespace std;
const int maxn=1011111;
int f[maxn];
int d[maxn];
int find(int x)
{
    if(f[x]==x)
        return x;

    int r=find(f[x]);
    d[x]=d[f[x]]+d[x];//当前x到她老祖宗的距离等于他现在的距离加上她爸爸到老祖宗的距离

    return f[x]=r; //把每个子节点合并到他的祖先
}


int fix(int x,int y,int z)
{
    int fx=find(x);
    int fy=find(y);
    if(fx==fy)
        return z==d[y]-d[x];
    f[fy]=fx;//把x当作y的爸爸；
    d[fy]=d[x]+z-d[y];
    return 1;
}





int main()
{
    int n,m;
    int lose=1;
    cin>>n>>m;
    for(int i=1;i<=n;i++)//初始化：先把所有点的爸爸都设为自己；
        f[i]=i;
    for(int i=1;i<=m;i++)
    {
        int x,y,z;
        cin>>x>>y>>z;

        if(lose)
            lose=fix(x,y,z);

    }

    if(lose)
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;
    return 0;
}