Wormholes(最短路)

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 52936 Accepted: 19732

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 backT 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

spfa

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <queue>#define inf 0x3f3f3f3fusing namespace std;int dis[505], vis[505], first[505], num[505], next[8005], u[8005], v[8005], w[8005], sum; //注意边的数目要开到二倍int spfa(int n);void add(int x, int y, int z){    u[sum] = x, v[sum] = y, w[sum] = z;    next[sum] = first[x];    first[x] = sum;    sum++;}int main(){    int f, x, y, z, n, m, k;    scanf("%d", &f);    while(f--)    {        scanf("%d %d %d", &n, &m, &k);        for(int i = 1; i <= n; i++)        {            num[i] = 0;            vis[i] = 0;            first[i] = -1;            dis[i] = inf;        }        sum = 1;        for(int i = 1; i <= m; i++)            {                scanf("%d %d %d", &x, &y, &z);                add(x, y, z);                add(y, x, z);            }        for(int i = 1; i <= k; i++)            {                scanf("%d %d %d", &x, &y, &z);                add(x, y, -z);            }        if(spfa(n))                printf("YES\n");        else                printf("NO\n");    }    return 0;}int spfa(int n){    int k;    queue<int>q;    dis[1] = 0;    vis[1] = 1;    num[1]++;    q.push(1);    while(q.size() > 0)        {            k = first[q.front()];            vis[q.front()] = 0;            q.pop();            while(k != -1)            {                if(dis[v[k]] > dis[u[k]] + w[k])                    {                        dis[v[k]] = dis[u[k]] + w[k];                        if(vis[v[k]] == 0)                            {                                q.push(v[k]);                                vis[v[k]] = 1;                                num[v[k]]++;                                if(num[v[k]] > n)                                    return 1;                            }                    }                k = next[k];            }        }    return 0;}

floyd

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#define inf 0x3f3f3f3fusing namespace std;int maps[550][550];int main(){    int f, n, m, w, x, y, z;    scanf("%d", &f);    while(f--)    {        bool flag = 0;        scanf("%d %d %d", &n , &m , &w);        memset(maps,0x3f3f3f3f,sizeof(maps));            for(int i=1;i<=n;i++)maps[i][i]=0;        for(int i = 1; i <= m; i++)            {                scanf("%d %d %d", &x, &y, &z);                if(z < maps[x][y])maps[x][y]=maps[y][x]=z;            }        for(int i = 1; i <= w; i++)            {                scanf("%d %d %d", &x, &y, &z);                maps[x][y] = min(maps[x][y], -z);            }        for(int k = 1; k <= n; k++)               {                for(int i = 1; i <= n; i++)                    {                    for(int j = 1; j <= n; j++)                           {                            int t = maps[i][k] + maps[k][j];                            if(maps[i][j] > t)                            maps[i][j] = t;                           }                    if(maps[i][i] < 0)                        flag = 1;                    }                    if(flag)                        break;               }        if(flag)                printf("YES\n");        else                printf("NO\n");    }    return 0;}