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AtCoder Beginner Contest 087 D - People on a Line

发布时间：2023/10/11 综合教程 12 老码农
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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.

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.

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).

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.

Sample Input 3

Copy
4 3

2 1 1

2 3 5

3 4 2

Sample Output 3

Copy
Yes

Sample Input 4

Copy
10 3

8 7 100

7 9 100

9 8 100

Sample Output 4

Copy
No

Sample Input 5

Copy
100 0

Sample Output 5

Copy
Yes

dfs，用邻接表标记每一个访问过的点，没访问的，更新dis（dis为当前子图中任意一点到其他点的距离），访问过看一下是否与dis匹配，不匹配就不满足。
代码：

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <iomanip>

using namespace std;

struct info

{

    int l,r,d;

}s[];

int n,m;

int flag = ;

int dis[],visited[];

int first[],nexti[];

void dfs(int t)

{

    if(!flag)return;

    int k = first[t];

    while(k != -)

    {

        if(!flag)return;

        if(!visited[s[k].r])

        {

            dis[s[k].r] = dis[s[k].l] + s[k].d;

            visited[s[k].r] = ;

            dfs(s[k].r);

        }

        else if(s[k].d != (dis[s[k].r] - dis[s[k].l]))

        {

            flag = ;

            return;

        }

        k = nexti[k];

    }

}

int main()

{

    scanf("%d%d",&n,&m);

    memset(first,-,sizeof(first));

    for(int i = ;i < m;i ++)

    {

        scanf("%d%d%d",&s[i].l,&s[i].r,&s[i].d);

        nexti[i] = first[s[i].l];

        first[s[i].l] = i;

        s[i + m].r = s[i].l;

        s[i + m].l = s[i].r;

        s[i + m].d = -s[i].d;

        nexti[i + m] = first[s[i + m].l];

        first[s[i + m].l] = i + m;

    }

    for(int i = ;i <= n;i ++)

    {

        if(!visited[i])

        {

            visited[i] = ;

            dfs(i);

        }

    }

    if(flag)printf("Yes");

    else printf("No");

}

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