Currency Exchange POJ - 1860
考察:spfa或者BF
本题边有多个属性值,所以需要多开几个数组
正确思路:
因为图是成环形的,最终会回到原点,如果值变小了最后队列会为empty,如果值变大了,队列会在i==s时,跳出循环
本题最好重新做过一遍
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 #include <algorithm> 6 using namespace std; 7 const int N = 110; 8 int n,m,s; 9 double v,g[N][N],d[N][N],dist[N]; 10 bool st[N]; 11 bool spfa(int s) 12 { 13 queue<int> q; 14 q.push(s); 15 dist[s] = v; 16 while(q.size()) 17 { 18 int t = q.front(); 19 st[t] = 0; 20 q.pop(); 21 for(int i=1;i<=n;i++) 22 { 23 if(dist[i]<(dist[t]-d[t][i])*g[t][i]) 24 { 25 dist[i] = (dist[t]-d[t][i])*g[t][i]; 26 if(i==s) return true; 27 if(!st[i]) q.push(i),st[i] = 1; 28 } 29 } 30 } 31 return false; 32 } 33 int main() 34 { 35 // freopen("in.txt","r",stdin); 36 scanf("%d%d%d%lf",&n,&m,&s,&v); 37 for(int i=1;i<=n;i++) g[i][i] =1; 38 for(int i=1;i<=m;i++) 39 { 40 int x,y; double toy1,toy2,tox1,tox2; 41 scanf("%d%d%lf%lf%lf%lf",&x,&y,&toy1,&toy2,&tox1,&tox2); 42 g[x][y] = toy1,g[y][x] = tox1; 43 d[x][y] = toy2,d[y][x] = tox2; 44 } 45 if(spfa(s)) puts("YES"); 46 else puts("NO"); 47 return 0; 48 }