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
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
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[400005]; int n,m; int flag = 1; int dis[100005],visited[100005]; int first[400005],nexti[400005]; void dfs(int t) { if(!flag)return; int k = first[t]; while(k != -1) { if(!flag)return; if(!visited[s[k].r]) { dis[s[k].r] = dis[s[k].l] + s[k].d; visited[s[k].r] = 1; dfs(s[k].r); } else if(s[k].d != (dis[s[k].r] - dis[s[k].l])) { flag = 0; return; } k = nexti[k]; } } int main() { scanf("%d%d",&n,&m); memset(first,-1,sizeof(first)); for(int i = 0;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 = 1;i <= n;i ++) { if(!visited[i]) { visited[i] = 1; dfs(i); } } if(flag)printf("Yes"); else printf("No"); }