POJ - 3259 Wormholes（最短路 Bellman-Ford算法）

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 57342		Accepted: 21388

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

Sample Output

NO
YES

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.

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 10000
#define INF 0x3f3f3f3f

using namespace std;

int dis[N];
int n,m,W,num;

struct node
{
    int s,e,t;
}e[N];

bool Bellman_Ford()
{
    int i,j,flag;   ///flag判斷負邊權
    for(i=0;i<n;i++)
    {
        dis[i] = INF;
    }
    dis[1] = 0;
    for(i=0;i<n;i++)
    {
        flag = 0;
        for(j=0;j<num;j++)
        {
            if(dis[e[j].e] > dis[e[j].s] + e[j].t)
            {
                dis[e[j].e] = dis[e[j].s] + e[j].t;
                flag = 1;
            }
        }
        if(!flag)   ///沒有鬆弛，一定沒有負邊
            return false;
    }
    for(i=0;i<num;i++)
    {
        if(dis[e[i].e] > dis[e[i].s] + e[i].t)
            return true;    ///存在負邊
    }
    return false;
}

int main()
{
    int T,i,j;
    scanf("%d",&T);
    while(T--)
    {
        num = 0;
        int u,v,w;
        scanf("%d %d %d",&n,&m,&W);
        for(i=0; i<m; i++)
        {
            scanf("%d %d %d",&u,&v,&w);
            e[num].s = u;
            e[num].e = v;
            e[num].t = w;
            num++;
            e[num].s = v;
            e[num].e = u;
            e[num].t = w;
            num++;
        }
        for(i=0;i<W;i++)
        {
            scanf("%d %d %d",&u,&v,&w);
            e[num].s = u;
            e[num].e = v;
            e[num].t = -w;
            num++;
        }
        if(Bellman_Ford())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

反思：

題目大意：約翰農夫發現農場有蟲洞，因爲他是個狂熱的時間旅行愛好者，所以打算皮一下，想試試能不能通過蟲洞回到起點看到自己。農場的路徑是雙向的，蟲洞是單向的負邊。第一行是數據組數，每組的第一行n,m,w分別是農場數、路徑數和蟲洞數，接下來m行是連接兩邊的點和邊的權值，再接下來w行是存在蟲洞的路徑。最短路問題，存在負權邊，所以選用Bellman-Ford算法。如果存在負權邊，則約翰可以穿越時空看到自己，輸出YES，否則就不可以，輸出NO。