poj Wormholes（Bellman_ford寻找负权环）

 Wormholes

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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.

思路：判断是否存在负权环，Bellman_ford和spfa皆可（注意路是双向的，虫洞是单向的）

Bellman_ford代码：

#include<stdio.h>#include<string.h>#define maxn 600+10#define maxv 6000+10const int inf=0x3f3f3f3f;int d[maxn];struct node{    int u,v,w;} E[maxv];int n,m,s,len;void Bellman_ford(){    for(int i=1; i<=n; i++)d[i]=inf;    int flag=1,cnt=0;    while(flag)    {        flag=0;        if(++cnt>n)        {            printf("YES\n");            return ;        }        for(int j=1; j<len; j++)        {            int x=E[j].u,y=E[j].v;            if(d[y]>d[x]+E[j].w)                d[y]=d[x]+E[j].w,flag=1;        }    }    printf("NO\n");}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d",&n,&m,&s);        int i,x,y,z;        len=1;        for(i=1; i<=m; i++)        {            scanf("%d%d%d",&x,&y,&z);            E[len].u=x,E[len].v=y,E[len].w=z,len++;            E[len].u=y,E[len].v=x,E[len].w=z,len++;        }        for(i=1; i<=s; i++)        {            scanf("%d%d%d",&x,&y,&z);            E[len].u=x,E[len].v=y,E[len].w=-z,len++;        }        Bellman_ford();    }    return 0;}

spfa代码：

#include<stdio.h>#include<string.h>#include<queue>using namespace std;#define maxn 500+10#define maxv 6000+10const int inf=0x3f3f3f3f;struct node{    int u,v,w;}E[maxv];int n,m,s,len;int d[maxn];int first[maxn],next[maxv];bool inq[maxn];int vis[maxn];bool spfa(int st){    memset(inq,0,sizeof(inq));    memset(vis,0,sizeof(vis));    for(int i=1;i<=n;i++)d[i]=inf;    d[st]=0,inq[st]=1;    queue<int>q;    q.push(st);    while(!q.empty())    {        int now=q.front();        q.pop();        inq[now]=0;        for(int i=first[now];i!=-1;i=next[i])        {            int x=E[i].v,y=E[i].w;            if(d[x]>d[now]+y)            {                d[x]=d[now]+y;                if(!inq[x])                {                    inq[x]=1;                    vis[x]++;                    q.push(x);                    if(vis[x]>n)                        return true;                }            }        }        if(d[st]<0)            return true;    }    return false;}void add_egde(int u,int v,int w){    E[len].u=u,E[len].v=v,E[len].w=w;    next[len]=first[u];    first[u]=len++;}void solve(){    for(int i=1;i<=n;i++)    {        if(spfa(i))        {            printf("YES\n");            return ;        }    }    printf("NO\n");}int main(){    int t;    scanf("%d",&t);    while(t--)    {        memset(first,-1,sizeof(first));        scanf("%d%d%d",&n,&m,&s);        int u,v,w,i;        len=1;        for(i=1;i<=m;i++)        {            scanf("%d%d%d",&u,&v,&w);            add_egde(u,v,w);            add_egde(v,u,w);        }        for(i=1;i<=s;i++)        {            scanf("%d%d%d",&u,&v,&w);            add_egde(u,v,-w);        }        solve();    }    return 0;}

ps：此题spfa比较慢，还是Bellman_ford比较好，但是测试数据好像比较弱，直接把起点看成1也能过 0.0