AtCoder Regular Contest 088 D - People on a Line（DFS+图论）

AtCoder Regular Contest 088 D - People on a Line（DFS+图论）

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.

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.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

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

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

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

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

用邻接表储存图，然后遍历，记录每个点之间距离，如果再次到达访问的点，发现距离和之前储存的不同则返回false

最后返回true；

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int N = 1e5+10;
int n,m;
int dis[N];
bool vis[N];
vector<pair<int,int> >E[N];
bool dfs(int u){
    vis[u] = 1;
    vector<pair<int,int> >::iterator it;
    for(it = E[u].begin(); it != E[u].end(); it++){
        int v = (*it).first;//注意这里的迭代器要加（*）
        int d = (*it).second;
        if(!vis[v]){
            dis[v] = dis[u] + d;
            if(!dfs(v))
                return false;
        }
        else if(dis[u]+d!=dis[v])
            return false;
    }
    return true;
}
int main(){
    cin >> n >> m;
    int u,v,w;
    int i;
    for(i = 1; i <= m; i++){
        cin >> u >> v >> w;
        E[u].push_back(make_pair(v,w));
        E[v].push_back(make_pair(u,-w));
    }
    for(i = 1; i <= n; i++){
        if(!vis[i]){
            if(!dfs(i)){
                cout << "No" << endl;
                return 0;
            }
        }
    }
    cout << "Yes" << endl;
    return 0;
}