POJ 3259 Wormholes (Bellman-Ford判断负环)

题目

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.

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题意

有一个人，他喜欢时间旅行，现在有一些虫洞，可以回到过去，现在有 n 个点，m 条边，代表现在可以走的通路，比如从 a 到 b 和从 b 到 a 需要花费 c 时间，现在在地上出现了 w 个虫洞，虫洞的意义就是你从 a 到 b 话费的时间是 −c (时间倒流,并且虫洞是单向的)，现在问你从某个点开始走，能回到从前。现在让你判断他能不能回到从前。

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思路

题目的数据给出了每个点的坐标，这样就可以构成一张图，用 Bellman−Ford 算法判断该图中是否存在负环，如果存在，输出“YES”,否则输出“NO”。题目数据比较水，没有考虑重边的情况。。。可以套模板

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代码

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<queue>#include<stack>#include<vector>#include<cmath>#include<set>#include<map>#include<cstdlib>#include<functional>#include<climits>#include<cctype>#include<iomanip>using namespace std;typedef long long ll;#define INF 0x3f3f3f3f#define mod 1e9+7#define clr(a,x) memset(a,x,sizeof(a))const double eps = 1e-6;const int MAX_N=10000;const int MAX_E=10000;int From[MAX_E],To[MAX_E],W[MAX_E];int dis[MAX_N],tot;void init(){    tot=0;}void add_edge(int u,int v,int d){    From[tot]=u;    To[tot]=v;    W[tot++]=d;}bool Bellman_ford(int s,int n){    clr(dis,0x3f);    dis[s]=0;    for(int k=0;k<n-1;k++)    {        bool relaxed=0;        for(int i=0;i<tot;i++)        {            int x=From[i],y=To[i];            if(dis[y]>dis[x]+W[i])            {                dis[y]=dis[x]+W[i];                relaxed=1;            }        }        if(!relaxed)            break;    }    for(int i=0;i<tot;i++)        if(dis[To[i]]>dis[From[i]]+W[i])            return 1;    return 0;}int main(){    int f,n,m,t,u,v,w;    cin>>f;    while(f--)    {        int num=0;        init();        cin>>n>>m>>t;//*************************************建图        for(int i=0;i<m;i++)        {            cin>>u>>v>>w;            add_edge(u,v,w);            add_edge(v,u,w);            num+=2; //记录边的个数        }                               for(int i=0;i<t;i++)        {            cin>>u>>v>>w;            add_edge(u,v,-w);        }//*************************************建图        bool ans=Bellman_ford(1,num);        if(ans)            cout<<"YES"<<endl;        else            cout<<"NO"<<endl;    }    return 0;}