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注意:在这个题目中,每个城市可以经过多次,且每条边也可能走多次 (我就是没有想到每条边还可能走多次。。。卡了一上午)这个要好好想想#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; const int maxn = 15; struct edge { int v,c,p,r,next; }edges[maxn]; int head[maxn]; int vispath[maxn]; //bool vism[maxn]; int n,m,e,ans; void addedge(int u,int v,int c,int p,int r) { edges[e].v = v; edges[e].c = c; edges[e].p = p; edges[e].r = r; edges[e].next = head[u]; head[u] = e++; } void dfs(int u,int cost) { if(cost > ans) return; if(u == n) { if(cost < ans) ans = cost; return; } for(int i=head[u];i!=-1;i=edges[i].next) { int v = edges[i].v; if(vispath[v]<=3) { vispath[v]++; if(vispath[edges[i].c]>0) dfs(v,cost+edges[i].p); else dfs(v,cost+edges[i].r); vispath[v]--; } } } int main() { int u,v,c,p,r; while(scanf("%d%d",&n,&m)!=EOF) { memset(head,-1,sizeof(head)); e = 0; for(int i=0;i<m;i++) { scanf("%d%d%d%d%d",&u,&v,&c,&p,&r); addedge(u,v,c,p,r); } ans = 99999999; memset(vispath,0,sizeof(vispath)); memset(vism,false,sizeof(vism)); vispath[1] = 1; dfs(1,0); if(ans < 99999999) printf("%d\n",ans); else printf("impossible\n"); } return 0; } Followed by: Post your reply here: |
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