AtCoder Regular Contest 090 D - People on a Line

D - People on a Line

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

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Sample Input 5

Copy

100 0

Sample Output 5

Copy

Yes

题意：给定有N个人，M条信息。每条信息：L，R，D，表示第R个人在第L个人的右边距离为D。求根据这M条信息判断安排是否成立。
题解：本来以为可以在线一条一条输入输入信息的同时判断是否合法，好像是不可以。答案说是构成图，然后DFS每一个节点。

 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<vector>
 #include<cstdio>
 using namespace std;
 const int maxn = 1e5 + ;
 typedef long long ll;
 vector<pair<int, int> > e[maxn];
 ll dis[maxn];
 bool vis[maxn];
 int n, m,flag=;

 void dfs(int x)
 {
     vis[x] = ;
     for (int i = ; i < e[x].size(); i++)
     {
         pair<int, int> cur=e[x][i];
         if (vis[cur.first]) {
             if (dis[cur.first] - dis[x] != cur.second) {
                 flag = ; break;
             }
         }
         else {
             dis[cur.first] = dis[x] + cur.second;
             dfs(cur.first);
         }
     }
 }

 int main()
 {
     cin >> n >> m;
     memset(vis, , sizeof(vis));
     for (int i = ; i < m; i++) {
         int l, r, d;
         cin >> l >> r >> d;
         e[l].push_back({ r,d });
         e[r].push_back({ l,-d });
     }
     flag = ;
     for (int i = ; i <n; i++) {
         if (!vis[i])
             dfs(i);
         if (flag) {
             break;
         }
     }
     if (flag) {
         cout << "No" << endl;
     }
     else
         cout << "Yes" << endl;
     return ;
 }