最短路练习6 /poj/3259 Wormholes 有负环

题目链接：http://poj.org/problem?id=3259

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 50025 Accepted: 18449

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.

题意：一种路是双向的，路的长度是正值；另一种路是单向的，路的长度是负值；  如果有负环输出YES；否则输出NO；

不同的路可能有相同的起点和终点：必须用邻接表（一种是数组模拟，一种是链表）

我的博客http://blog.csdn.net/xiangaccepted/article/details/70172876有数组模拟的模拟过程（不会的可以看一下）

思路：spfa判断负环；

AC代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<vector>#include<map>#include<string>#define LL long long#define eps 1e-8using namespace std;const int mod = 1e7+7;const int inf = 0x3f3f3f3f;const int maxn = 1e6 +10;int t,n,m,w,a,b,c,cnt,ans;struct edge{    int l,r,num,next;//起始点l，终点r，路的长度num，起始点能到达的上一个地方的编号；}e[10000];int head[550];//head[k]记录地点k能去的最后一个地方的编号；int vis[550];//标记   是否在队列中int dis[550];//最短距离int sum[550];//进入队列的次数；void add(int ll,int rr,int numm){    e[cnt].l=ll,e[cnt].r=rr,e[cnt].num=numm;    e[cnt].next=head[ll];//e[cnt].next存上一个ll能到达的地点的编号；    head[ll]=cnt++;//当前的head[ll]已经被用过了，再次更新编号；}void init()//初始化{    memset(head,-1,sizeof(head));    memset(e,0,sizeof(e));    cnt=0;    ans=0;}queue<int>q;int spfa(int x){    while(!q.empty())        q.pop();    memset(dis,0x3f,sizeof(dis));//初始化    memset(vis,0,sizeof(vis));    memset(sum,0,sizeof(sum));    q.push(x);    dis[x]=0;    vis[x]=1;    sum[x]=1;    while(!q.empty())//spfa模板    {        int tem=q.front();        q.pop();        vis[tem]=0;        for(int i=head[tem];i!=-1;i=e[i].next)        {            if(dis[e[i].r]>dis[tem]+e[i].num)            {                dis[e[i].r]=dis[tem]+e[i].num;                if(!vis[e[i].r])                {                    vis[e[i].r]=1;                    sum[e[i].r]++;                    if(sum[e[i].r]>n) return 1;//如果sum[e[i].r]>n说明这个点进入队列次数太多了，有负环，返回1；                    q.push(e[i].r);                }            }        }    }    return 0;}int main(){    scanf("%d",&t);    while(t--)    {        init();        scanf("%d%d%d",&n,&m,&w);        for(int i=0;i<m;i++)        {            scanf("%d%d%d",&a,&b,&c);            add(a,b,c);            add(b,a,c);//双向的        }        for(int i=0;i<w;i++)        {            scanf("%d%d%d",&a,&b,&c);            add(a,b,c*(-1));//单向，道路为负值        }        for(int i=1;i<=n;i++)//判断每个点是否有负环        {            if(spfa(i))            {                ans=1;                break;            }        }        if(ans)            printf("YES\n");        else            printf("NO\n");    }}