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POJ – 3259 Wormholes (图论/floyd判断负圈)

编程语言 karry_zzj 20℃ 0评论
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问题描述


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

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

Sample Output

NO
YES

1.分析:

代码如下:

#include
#include
using namespace std;
const int maxn = 500+10; 
const int maxe = 5200+10;
struct edge{
    int from, to, cost;
}es[maxe];
int F;
int V,E;
int d[maxn];

/*判断是否有负圈*/
bool find_negative_loop()
{
    memset(d, 0, sizeof(d));
    for(int i=0; ifor(int j=0; jif(d[e.to] > d[e.from] + e.cost)
            {
                d[e.to] = d[e.from] + e.cost;
                if(i == V-1) return true;
            }
        }
    }
    return false;
}

int main()
{
    scanf("%d",&F);
    while(F--)
    {
        int N,M,W;
        scanf("%d%d%d",&N,&M,&W);
        V = N;
        E = 0; 
        int s,e,t;
        for(int i=0; iscanf("%d%d%d",&s,&e,&t);
            es[E].from = s;
            es[E].to = e;
            es[E].cost = t;
            E++;
            es[E].to = s;
            es[E].from = e;
            es[E].cost = t;
            E++;
        }   
        for(int i=0; iscanf("%d%d%d",&s,&e,&t);
            es[E].from = s;
            es[E].to = e;
            es[E].cost = -t;
            E++;
        }
        if(find_negative_loop()) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
} 

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