poj 3259 Wormholes spfa算法
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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 25582 | Accepted: 9186 |
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 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
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 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.
spaf是bellman算法的一个优化，每次松弛的不是全图的点，而是与最短距离有变化的点相连的那些点，题目用到一个队列，每个点如果最短路径更新了，就扔到队列里，然后从队列中拿出点来松弛与其相连的点，如果一个点入队列的次数达到n此，说明图中有负圈
注意两个点之间可能有多条路径，选择权值较小的那条边留下
#include
#include
#include
#include
using namespace std;
int map[501][501];
int dis[501];
int n, m, w;
int s, e, t;
bool spfa()
{
bool flag[501] = {0};
int count[501] = {0};
queue q;
q.push(s);
dis[s] = 0;
int curr;
int i;
while(!q.empty())
{
curr = q.front();
q.pop();
for(i = 1; i <= n; i++)
{
if(map[curr][i] < 100000)
{
if(dis[i] > map[curr][i] + dis[curr] )
{
dis[i] = map[curr][i] + dis[curr];
if(flag[i] == 0)
q.push(i);
count[i] ++ ;
flag[i] = 1;
if(count[i] >= n)
return 0;
}
}
}
flag[curr] = 0;
}
return 1;
}
int main()
{
int f;
scanf("%d", &f);
while(f--)
{
memset(dis,63, sizeof(dis));
memset(map, 127, sizeof(map));
scanf("%d %d %d", &n, &m, &w);
int i;
for(i = 0; i < m; i++)
{
scanf("%d %d %d", &s, &e, &t);
map[s][e] = map[s][e] > t? t : map[s][e];
map[e][s] = map[e][s] > t? t : map[e][s];
}
for(i = 0; i < w; i++)
{
scanf("%d %d %d", &s, &e, &t);
map[s][e] = -t;
}
if(spfa())
printf("NO\n");
else
printf("YES\n");
}
return 0;
}