POJ3259 解题报告

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 52092 Accepted: 19388

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

USACO 2006 December Gold

好的，说实话，这一题其实相当简单，但是坏就坏在我瞎了眼，有写错了变量名。把es[i].cost写成了es[i].to。发现了之后相当生气。。

这一题的题意是问FJ能不能在他出发前的时间回到起点，想要达到这个目的很显然是需要图中有负环的，那么自然就想到了用bellman算法

#include<iostream>#include<algorithm>#include<cstdio>using namespace std;struct edge{    int from;    int to;    int cost;};const int INF=2e9;const int maxn=500+50;const int maxm=2500+20;const int maxw=200+20;const int maxe=7000;edge es[maxe];int d[maxn];int F,N,M,W,E;void bellman_ford(int s);int main(){    cin>>F;    while(F--)    {        cin>>N>>M>>W;        E=2*M+W;        int i;        for(i=0;i<2*M;i+=2)        {            int s,e,t;            cin>>s>>e>>t;            es[i].from=s; es[i].to=e; es[i].cost=t;            es[i+1].from=e; es[i+1].to=s; es[i+1].cost=t;        }        for(int j=0;j<W;j++)        {            int s,e,t;            cin>>s>>e>>t;            es[i+j].from=s; es[i+j].to=e; es[i+j].cost=-1*t;        }        //cout<<E<<endl;        bellman_ford(1);    }    return 0;}void bellman_ford(int s){    fill(d,d+maxn,INF);    d[s]=0;    int loop=0;    for(int j=0;j<N;j++)    {        loop++;        bool update=false;        for(int i=0;i<E;i++)        {            edge e=es[i];            if(d[e.from]!=INF&&d[e.to]>d[e.from]+e.cost)            {                d[e.to]=d[e.from]+e.cost;                update=true;            }        }        if(!update) break;        if(loop==N)        {            cout<<"YES\n";            return ;        }    }    cout<<"NO\n";}