poj 3259 bellman最短路判断有无负权回路

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 36717 Accepted: 13438

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 toF (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 toE 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.

Source

70ms

#include<iostream>  //79ms#include<cstdio>#include<cstring>#include<cmath>#define INF 10000000using namespace std;struct node{    int u,v,w;} edge[5500];int low[5500];int n,m,z;int num=0;int Bellman(){    for(int i=0; i<=n; i++)        low[i]=INF;    for(int i=0; i<n-1; i++)    {        int flag=0;        for(int j=0; j<num; j++)        {            if(low[edge[j].u]+edge[j].w<low[edge[j].v])            {                low[edge[j].v]=low[edge[j].u]+edge[j].w;                flag=1;            }        }        if(flag==0)  //存在负权回路            break;    }    for(int j=0; j<num; j++)   //判断负权回路    {        if(low[edge[j].u]+edge[j].w<low[edge[j].v])            return 1;    }    return 0;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&m,&z);        int a,b,c;        num=0;        for(int i=1; i<=m; i++)        {            scanf("%d%d%d",&a,&b,&c);            edge[num].u=a;            edge[num].v=b;            edge[num++].w=c;            edge[num].u=b;            edge[num].v=a;            edge[num++].w=c;        }        for(int i=1; i<=z; i++)        {            scanf("%d%d%d",&a,&b,&c);            edge[num].u=a;            edge[num].v=b;            edge[num++].w=-c;        }        if(Bellman())            printf("YES\n");        else            printf("NO\n");    }}

700ms

#include<iostream>  //挨个点遍历#include<cstdio>#include<cstring>#include<cmath>#define INF 0x3f3f3f3fusing namespace std;struct node{    int u,v,w;} edge[5500];int low[550];int n,m,z;int num=0;int Bellman(int u0){    for(int i=0; i<=n; i++)        low[i]=INF;    low[u0]=0;    for(int i=0; i<n; i++)  //递推n次,让其构成环来判断    {        int flag=0;        for(int j=0; j<num; j++)        {            if(low[edge[j].u]!=INF&&low[edge[j].u]+edge[j].w<low[edge[j].v])            {                low[edge[j].v]=low[edge[j].u]+edge[j].w;                flag=1;            }        }        if(flag==0)  //存在负权回路(减少时间)            break;    }    if(low[u0]<0)return 1;    return 0;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&m,&z);        int a,b,c;        num=0;        for(int i=1; i<=m; i++)        {            scanf("%d%d%d",&a,&b,&c);            edge[num].u=a;            edge[num].v=b;            edge[num++].w=c;            edge[num].u=b;            edge[num].v=a;            edge[num++].w=c;        }        for(int i=1; i<=z; i++)        {            scanf("%d%d%d",&a,&b,&c);            edge[num].u=a;            edge[num].v=b;            edge[num++].w=-c;        }        int biao=1;        for(int i=1; i<=n; i++)        {            if(Bellman(i))            {                printf("YES\n");                biao=0;                break;            }        }        if(biao)            printf("NO\n");    }}/*780ms#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define INF 0x3f3f3f3fusing namespace std;struct node{    int u,v,w;} edge[5500];int low[5500];int n,m,z;int num=0;int Bellman(int u0){    for(int i=0; i<=n; i++)                     low[i]=INF;    low[u0]=0;                         //初始化    for(int i=0; i<n-1; i++)              //n-1次    {        int flag=0;        for(int j=0; j<num; j++)        {            if(low[edge[j].u]!=INF&&low[edge[j].u]+edge[j].w<low[edge[j].v])   //不同点            {                                          //存在low[edge[j].u]!=INF,就必须有low[u0]=0;初始化                low[edge[j].v]=low[edge[j].u]+edge[j].w;                flag=1;            }        }        if(flag==0)  //存在负权回路            break;    }    for(int j=0; j<num; j++)    {        if(low[edge[j].u]!=INF&&low[edge[j].u]+edge[j].w<low[edge[j].v])            return 1;    }    return 0;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&m,&z);        int a,b,c;        num=0;        for(int i=1; i<=m; i++)        {            scanf("%d%d%d",&a,&b,&c);            edge[num].u=a;            edge[num].v=b;            edge[num++].w=c;            edge[num].u=b;            edge[num].v=a;            edge[num++].w=c;        }        for(int i=1; i<=z; i++)        {            scanf("%d%d%d",&a,&b,&c);            edge[num].u=a;            edge[num].v=b;            edge[num++].w=-c;        }        int biao=1;        for(int i=1; i<=n; i++)        {            if(Bellman(i))            {                printf("YES\n");                biao=0;                break;            }        }        if(biao)            printf("NO\n");    }}*/