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
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
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。
1 #include<iostream> 2 #include<algorithm> 3 #include<cstring> 4 #include<vector> 5 #include<cstdio> 6 using namespace std; 7 const int maxn = 1e5 + 100; 8 typedef long long ll; 9 vector<pair<int, int> > e[maxn]; 10 ll dis[maxn]; 11 bool vis[maxn]; 12 int n, m,flag=0; 13 14 void dfs(int x) 15 { 16 vis[x] = 1; 17 for (int i = 0; i < e[x].size(); i++) 18 { 19 pair<int, int> cur=e[x][i]; 20 if (vis[cur.first]) { 21 if (dis[cur.first] - dis[x] != cur.second) { 22 flag = 1; break; 23 } 24 } 25 else { 26 dis[cur.first] = dis[x] + cur.second; 27 dfs(cur.first); 28 } 29 } 30 } 31 32 int main() 33 { 34 cin >> n >> m; 35 memset(vis, 0, sizeof(vis)); 36 for (int i = 0; i < m; i++) { 37 int l, r, d; 38 cin >> l >> r >> d; 39 e[l].push_back({ r,d }); 40 e[r].push_back({ l,-d }); 41 } 42 flag = 0; 43 for (int i = 0; i <n; i++) { 44 if (!vis[i]) 45 dfs(i); 46 if (flag) { 47 break; 48 } 49 } 50 if (flag) { 51 cout << "No" << endl; 52 } 53 else 54 cout << "Yes" << endl; 55 return 0; 56 }