POJ 3259 Wormholes（最短路，判断有没有负环回路）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24249		Accepted: 8652

Description

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.
Output

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).
Sample Input

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

Sample Output

NO
YES

Hint

For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.
Source

USACO 2006 December Gold

这题就是判断存不存在负环回路。
前M条是双向边，后面的W是单向的负边。

为了防止出现不连通，
增加一个结点作为起点。
起点到所有点的长度为0

bellman_ford算法：

/*
* POJ 3259
* 判断图中是否存在负环回路。
* 为了防止图不连通的情况，增加一个点作为起点，这个点和其余的点都相连。
*/

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
/*
* 单源最短路bellman_ford算法，复杂度O(VE)
* 可以处理负边权图。
* 可以判断是否存在负环回路。返回true,当且仅当图中不包含从源点可达的负权回路
* vector<Edge>E;先E.clear()初始化，然后加入所有边
* 点的编号从1开始(从0开始简单修改就可以了)
*/
const int INF=0x3f3f3f3f;
const int MAXN=550;
int dist[MAXN];
struct Edge
{
int u,v;
int cost;
Edge(int _u=0,int _v=0,int _cost=0):u(_u),v(_v),cost(_cost){}
};
vector<Edge>E;
bool bellman_ford(int start,int n)//点的编号从1开始
{
for(int i=1;i<=n;i++)dist[i]=INF;
dist[start]=0;
for(int i=1;i<n;i++)//最多做n-1次
{
bool flag=false;
for(int j=0;j<E.size();j++)
{
int u=E[j].u;
int v=E[j].v;
int cost=E[j].cost;
if(dist[v]>dist[u]+cost)
{
dist[v]=dist[u]+cost;
flag=true;
}
}
if(!flag)return true;//没有负环回路
}
for(int j=0;j<E.size();j++)
if(dist[E[j].v]>dist[E[j].u]+E[j].cost)
return false;//有负环回路
return true;//没有负环回路
}

int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
int T;
int N,M,W;
int a,b,c;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&N,&M,&W);
E.clear();
while(M--)
{
scanf("%d%d%d",&a,&b,&c);
E.push_back(Edge(a,b,c));
E.push_back(Edge(b,a,c));
}
while(W--)
{
scanf("%d%d%d",&a,&b,&c);
E.push_back(Edge(a,b,-c));
}
for(int i=1;i<=N;i++)
E.push_back(Edge(N+1,i,0));
if(!bellman_ford(N+1,N+1))printf("YES\n");
else printf("NO\n");
}
return 0;
}

SPFA算法：

//============================================================================
// Name        : POJ.cpp
// Author      :
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
/*
* 单源最短路SPFA
* 时间复杂度 0(kE)
* 这个是队列实现，有时候改成栈实现会更加快，很容易修改
* 这个复杂度是不定的
*/
const int MAXN=1010;
const int INF=0x3f3f3f3f;
struct Edge
{
int v;
int cost;
Edge(int _v=0,int _cost=0):v(_v),cost(_cost){}
};
vector<Edge>E[MAXN];
void addedge(int u,int v,int w)
{
E[u].push_back(Edge(v,w));
}
bool vis[MAXN];
int cnt[MAXN];
int dist[MAXN];
bool SPFA(int start,int n)
{
memset(vis,false,sizeof(vis));
for(int i=1;i<=n;i++)dist[i]=INF;
dist[start]=0;
vis[start]=true;
queue<int>que;
while(!que.empty())que.pop();
que.push(start);
memset(cnt,0,sizeof(cnt));
cnt[start]=1;
while(!que.empty())
{
int u=que.front();
que.pop();
vis[u]=false;
for(int i=0;i<E[u].size();i++)
{
int v=E[u][i].v;
if(dist[v]>dist[u]+E[u][i].cost)
{
dist[v]=dist[u]+E[u][i].cost;
if(!vis[v])
{
vis[v]=true;
que.push(v);
if(++cnt[v]>n)return false;
//有负环回路
}
}
}
}
return true;
}
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
int T;
int N,M,W;
int a,b,c;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&N,&M,&W);
for(int i=1;i<=N+1;i++)E[i].clear();
while(M--)
{
scanf("%d%d%d",&a,&b,&c);
addedge(a,b,c);
addedge(b,a,c);
}
while(W--)
{
scanf("%d%d%d",&a,&b,&c);
addedge(a,b,-c);
}
for(int i=1;i<=N;i++)
addedge(N+1,i,0);
if(!SPFA(N+1,N+1))printf("YES\n");
else printf("NO\n");
}
return 0;
}