POJ Wormholes 3259(最短路)

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 43742 Accepted: 16061

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 Tseconds.

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

上代码

Bellman_Ford

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>#include <queue>#include <iostream>using namespace std;#define inf 0x3f3f3f3fint top,n;int dist[1000];struct node{    int u,v,w;}q[6000];int bellman(){    int i,j;    for(i=1;i<=n;i++)    {        dist[i]=inf;    }    dist[1]=0;    for(i=1;i<n;i++)    {        for(j=0;j<top;j++)        {            int u=q[j].u;            int v=q[j].v;            if( dist[v]>dist[u]+q[j].w)            {                dist[v]=dist[u]+q[j].w;            }        }    }    for(i=0;i<top;i++)    {            int u=q[i].u;            int v=q[i].v;            if( dist[v]>dist[u]+q[i].w)            {                return 0;            }    }    return 1;}int main(){    int m,i,j,l,k;    int t;    int a,b,c;    cin>>t;    while(t--)    {        top=0;        cin>>n>>m>>l;        while(m--)        {            cin>>a>>b>>c;            q[top].u=a;            q[top].v=b;            q[top++].w=c;            q[top].u=b;            q[top].v=a;            q[top++].w=c;        }        while(l--)        {            cin>>a>>b>>c;            q[top].u=a;            q[top].v=b;            q[top++].w=-c;        }       if(!bellman())       {           printf("YES\n");       }       else       {           printf("NO\n");       }    }    return 0;}

SPFA

#include <queue>#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;#define inf 1<<26struct node{    int v,w;    int next;}q[6000];int head[6000],dist[6000],vis[6000],c[6000];int top,n;void chu(){    int i;    top=0;    for(i=0;i<=n;i++)    {        head[i]=-1;        vis[i]=0;        c[i]=0;    }}void add(int u,int v,int w){    q[top].v=v;    q[top].w=w;    q[top].next=head[u];    head[u]=top++;}void spfa(){    int i,j;    for(i=1;i<=n;i++)    {        dist[i]=inf;    }    vis[1]=1;    dist[1]=0;    c[1]++;    queue<int >Q;    Q.push(1);    while(!Q.empty())    {        int x=Q.front();        vis[x]=0;        Q.pop();        for(i=head[x];i!=-1;i=q[i].next)        {            int v=q[i].v;            int w=q[i].w;            if(dist[v]>dist[x]+w)            {                dist[v]=dist[x]+w;                if(!vis[v])                {                    vis[v]=1;                    c[v]++;                    if(c[v]>=n)                    {                        cout<<"YES\n";                        return ;                    }                    Q.push(v);                }            }        }    }    cout<<"NO\n";}int main(){    int t,m,i,j,l;    cin>>t;    int a,b,c;    while(t--)    {        cin>>n>>m>>l;        chu();        while(m--)        {            cin>>a>>b>>c;            add(a,b,c);            add(b,a,c);        }        while(l--)        {            cin>>a>>b>>c;            add(a,b,-c);        }        spfa();    }    return 0;}