AtCoder Beginner Contest 087 D - People on a Line

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 x_i. For every i, x_i 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 (L_i,R_i,D_i). This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_Ri−x_Li=D_i 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,…,x_N) that is consistent with the given pieces of information.

Constraints

• 1≤N≤100 000
• 0≤M≤200 000
• 1≤L_i,R_i≤N (1≤i≤M)
• 0≤D_i≤10 000 (1≤i≤M)
• L_i≠R_i (1≤i≤M)
• If i≠j, then (L_i,R_i)≠(L_j,R_j) and (L_i,R_i)≠(R_j,L_j).
• D_i 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,…,x_N) 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

---

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");
}