POJ 3259 Wormholes （判负环）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 46123		Accepted: 17033

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

题目大意：
FJ有n块农场，编号为1到n，这n块农场由m条道路和w个虫洞连接，没条道路是双向的，权值为t1，表示经过每条道路需要花费t1个时间单位，每个虫洞是单向的，权值为t2，经过每个虫洞可以让你回到t2个时间单位之前（说白了就是时光倒流）；现在问你，FJ想从1号农场开始，经过若干农场后，在其出发之前的某一时刻回到1号农场。现在问你能否实现。
分析：
我们把虫洞上的权值看成负的，这样问题就变成了问你这块农场中是否存在负环的问题了。

　

单纯spfa跑最短路，判一个点入队超过n次，就有负环，较慢。

这里写一种较快的，dfs版的spfa。详见代码。

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define R register
using namespace std;
inline int read(){
    R int x=0;bool f=1;
    R char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=0;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return f?x:-x;
}
const int N=1e5+10;
int T,n,m1,m2;
struct node{
    int v,w,next;
}e[N<<1];
int tot,head[N],dis[N];
bool can,flag[N];
void add(int x,int y,int z){
    e[++tot].v=y;e[tot].w=z;e[tot].next=head[x];head[x]=tot;
}
void spfa(int x){
    flag[x]=1;
    for(int i=head[x];i;i=e[i].next){
        int v=e[i].v,w=e[i].w;
        if(dis[v]>dis[x]+w){
            if(flag[v]||can){can=1;break;}
            dis[v]=dis[x]+w;
            spfa(v);
        }
    }
    flag[x]=0;
}
void Cl(){
    can=0;tot=0;
    memset(dis,0,sizeof dis);//注意不是inf 
    memset(head,0,sizeof head);
    memset(flag,0,sizeof flag);
}
int main(){
    T=read();
    while(T--){
        Cl();
        n=read();m1=read();m2=read();
        for(int i=1,x,y,z;i<=m1;i++){
            x=read();y=read();z=read();
            add(x,y,z);
            add(y,x,z);
        }
        for(int i=1,x,y,z;i<=m2;i++){
            x=read();y=read();z=read();
            add(x,y,-z);
        }
        for(int i=1;i<=n;i++){
            spfa(i);
            if(can) break;
        }
        puts(can?"YES":"NO");
    }
    return 0;
}