标签:stdin hit 选中 its script 标记 break str with
3 2 30 20 10 0 6 2 6 0 3 2 3 0 2 2 1 1 0 2 2 0 0 0
1 3 1 2
最后从1到n扫描ans数组就可以保证答案是升序。
參考代码:
#include<stack> #include<queue> #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #pragma commment(linker,"/STACK: 102400000 102400000") using namespace std; const double eps=1e-6; const int INF=0x3f3f3f3f; const int MAXN=20; int n,m,mincost[MAXN],node[MAXN],edge[MAXN][MAXN]; bool vis[MAXN],used[MAXN],ans[MAXN]; double temp; double prim() { memset(mincost,0,sizeof(mincost)); memset(used,false,sizeof(used)); int s; for(int i=1; i<=n; i++) if(vis[i]) { s=i; break; } for(int i=1; i<=n; i++) { mincost[i]=edge[s][i]; used[i]=false; } used[s]=true; int res=0; int nodevalue=node[s]; for(int j=1; j<n; j++) { int v=-1; for(int i=1; i<=n; i++) if(vis[i]&&!used[i]&&(v==-1||mincost[v]>mincost[i]))//vis[i]表示第i个点有没有被选中 v=i; res+=mincost[v]; nodevalue+=node[v]; used[v]=true; for(int i=1; i<=n; i++) if(vis[i]&&!used[i]&&mincost[i]>edge[v][i]) mincost[i]=edge[v][i]; } return (res+0.0)/nodevalue; } void dfs(int pos,int num) { if(num>m) return ; if(pos==n+1) { if(num!=m) return ; double tans=prim(); if(tans<temp) { temp=tans; memcpy(ans,vis,sizeof(ans)); } return ; } vis[pos]=true;//选择当前点 dfs(pos+1,num+1); vis[pos]=false;//不选择当前点,消除标记 dfs(pos+1,num); } int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif // ONLINE_JUDGE while(scanf("%d%d",&n,&m)) { if(n==0&&m==0) break; for(int i=1; i<=n; i++) scanf("%d",&node[i]); for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) scanf("%d",&edge[i][j]); temp=INF; dfs(1,0); bool flag=false; for(int i=1; i<=n; i++) { if(ans[i]) { if(flag) printf(" "); else flag=true; printf("%d",i); } } printf("\n"); } return 0; }
HDU 2489 Minimal Ratio Tree(prim+DFS)
标签:stdin hit 选中 its script 标记 break str with
原文地址:http://www.cnblogs.com/zhchoutai/p/7084603.html