poj3259 Wormholes

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 29805		Accepted: 10779

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

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

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

 

解题：Bellman-Ford 算法模板题。存在负环即可以看见自己，虫洞？你懂的！

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <climits>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;
const int INF = INT_MAX>>1;
int cnt,dis[600];
struct ARC {
    int u,v,w;
} arc[6000];
bool bf(int n) {
    int i,j;
    for(i = 2; i <= n; i++)
        dis[i] = INF;
    dis[1] = 0;
    for(i = 1; i < n; i++) {
        for(j = 0; j < cnt; j++) {
            if(dis[arc[j].v] > dis[arc[j].u]+arc[j].w) {
                dis[arc[j].v] = dis[arc[j].u] + arc[j].w;
            }
        }
    }
    for(j = 0; j < cnt; j++) {
        if(dis[arc[j].v] > dis[arc[j].u]+arc[j].w) return false;
    }
    return true;
}
int main() {
    int kase,n,m,k,u,v,w,i;
    scanf("%d",&kase);
    while(kase--) {
        scanf("%d %d %d",&n,&m,&k);
        cnt = 0;
        while(m--) {
            scanf("%d %d %d",&u,&v,&w);
            arc[cnt++] = (ARC) {u,v,w};
            arc[cnt++] = (ARC) {v,u,w};
        }
        while(k--) {
            scanf("%d %d %d",&u,&v,&w);
            arc[cnt++] = (ARC) {u,v,-w};
        }
        bf(n)?puts("NO"):puts("YES");
    }
    return 0;
}

 解法2：SPFA算法，邻接表存储图。不过此题貌似效果不明显，还没裸的Bellman-Ford快！

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <climits>
#include <algorithm>
#include <cmath>
#include <queue>
#define LL long long
using namespace std;
const int maxn = 510;
const int INF = INT_MAX>>2;
int n;
struct arc {
    int w,to;
};
vector<arc>e[maxn];
bool used[maxn];
int num[maxn],d[maxn];
bool spfa(int x) {
    int i,j,temp;
    memset(num,0,sizeof(num));
    memset(used,false,sizeof(used));
    for(i = 1; i <= n; i++)
        d[i] = INF;
    d[x] = 0;
    queue<int>q;
    while(!q.empty()) q.pop();
    used[x] = true;
    num[x]++;
    q.push(x);
    while(!q.empty()) {
        temp = q.front();
        q.pop();
        used[temp] = false;
        for(i = 0; i < e[temp].size(); i++) {
            j = e[temp][i].to;
            if(d[j] > e[temp][i].w+d[temp]) {
                d[j] = e[temp][i].w + d[temp];
                if(!used[j]) {
                    num[j]++;
                    used[j] = true;
                    if(num[j] >= n) return true;
                    q.push(j);
                }
            }
        }
    }
    return false;
}
int main() {
    int kase,i,u,v,w,m,k;
    scanf("%d",&kase);
    while(kase--) {
        for(i = 0; i < maxn; i++)
            e[i].clear();
        scanf("%d %d %d",&n,&m,&k);
        while(m--) {
            scanf("%d %d %d",&u,&v,&w);
            e[u].push_back((arc) {w,v});
            e[v].push_back((arc {w,u}));
        }
        while(k--) {
            scanf("%d %d %d",&u,&v,&w);
            e[u].push_back((arc) {-w,v});
        }
        spfa(1)?puts("YES"):puts("NO");
    }
}