AtCoder Beginner Contest 087 D People on a Line(DFS)

题意

给出n个点，m组关系L,R,D，L在R的左边距离D，判断是否存在n个人的位置满足m组关系

分析

Consider the following directed graph G:

There are N vertices numbered 1,2,...,N.

For each i, there are an edge from vertex Li to vertex Ri with weight Di, and an edge from vertex

Ri to vertex Li with weight −Di.

The problem asks whether we can assign an integer xv to each vertex v in G, such that for each edgefrom u to v with cost d, xv − xu = d holds. (Clearly, we can ignore the condition 0 ≤ xi ≤ 109.)

We handle each connected component in G independently. For each connected component, we choosean arbitary vertex v and assume that xv = 0. By running a dfs from v, we can uniquly determine thevalues of xi in this component. After that, we should check if the conditions are actually satisfied.

思路是建图+DFS

建图：对于L,R,D，L->D设置长度为D，R->L设置长度为-D

DFS：遍历每个联通块，对于一个点，若它被访问过，利用dis判断是否合法，若未被访问过，标记并dfs

#include <bits/stdc++.h>
using namespace std;

#define ll long long

int n,m;
int L,R,D;
vector<pair<int,int> >mp[100100];
int dis[100100];
bool vis[100100];
int flag;

void dfs(int u,int pre)
{
	if(!flag) return;
	 int sz=mp[u].size();
	 pair<int,int>tmp;
	 for(int i=0;i<sz;++i)
	 {
	 	tmp=mp[u][i];
	 	if(vis[tmp.first])
	 	{
	 		if(dis[u]+tmp.second!=dis[tmp.first])
	 		{
	 			flag=0;return;
	 		}
	 	}
	 	else
	 	{
	 		vis[tmp.first]=1;
	 		dis[tmp.first]=dis[u]+tmp.second;
	 		dfs(tmp.first,u);
	 	}
	 }
}

int main(int argc, char const *argv[])
{
	/* code */
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;++i)
	{
		scanf("%d %d %d",&L,&R,&D);
		mp[R].push_back(make_pair(L,D));
		mp[L].push_back(make_pair(R,-D));
	}
	flag=1;
	for(int i=1;i<=n;++i) if(!vis[i])
	{
		vis[i]=1;
		dfs(i,-1);
	}
	if(flag) puts("Yes");else puts("No");
	return 0;
}