标签:blog os io strong 文件 for ar div
题意:给定完全无向图,求其中m个子节点,要求Sum(edge)/Sum(node)最小。
思路:由于N很小,枚举所有可能的子节点可能情况,然后求MST,memset()在POJ里面需要memory头文件。
#include <iostream> #include <vector> #include <map> #include <cmath> #include <memory> #include <algorithm> #include <cstdio> #include <cstdlib> using namespace std; const int MAXN = 100; const int INF = 1<<30; #define CLR(x,y) memset(x,y,sizeof(x)) #define MIN(m,v) (m)<(v)?(m):(v) #define MAX(m,v) (m)>(v)?(m):(v) #define ABS(x) ((x)>0?(x):-(x)) #define rep(i,x,y) for(i=x;i<y;++i) int n,m,k; double ans = 0; int select[MAXN]; int g[MAXN][MAXN]; int val[MAXN]; int visit[MAXN]; double dist[MAXN]; int ind[MAXN]; double Prim() { int i,j,tmp,mark_i; int mark_min; int sum_node = 0; int sum_edge = 0; for(i = 0; i < n; ++i) { dist[i] = INF; visit[i] = 0; } for(i = 0; i < n; ++i) if(select[i]>0) { dist[i] = 0; break; } int cnt = 0; for(i = 0; i < n; ++i,++cnt) { if(cnt>=m) break; mark_min = INF; for(j = 0; j < n; ++j) { if(select[j]>0 && !visit[j] && mark_min>dist[j]) { mark_min = dist[j]; mark_i = j; } } visit[mark_i] = 1; sum_edge += dist[mark_i]; for( j = 0; j < n; ++j) { if(visit[j]==0 && select[j]>0 && dist[j] > g[mark_i][j]) dist[j] = g[mark_i][j]; } } for( i = 0; i < n; ++i) if(select[i] > 0 ) sum_node += val[i]; return double(sum_edge)/sum_node; } int DFS(int cur, int deep) { double res = INF; select[cur] = 1; if(deep < m) { for(int i = cur+1; i < n; ++i) { DFS(i,deep+1); } } if(deep == m) { res = Prim(); if(res < ans) { ans = res; for(int i = 0; i < n; ++i) ind[i] = select[i]; } } select[cur] = 0; return 0; } int Solve() { while(scanf("%d%d",&n,&m)!=EOF) { if(n == 0 && m == 0) break; for(int i = 0 ; i < n; ++i) scanf("%d",&val[i]); for(int i = 0; i < n; ++i) for(int j = 0; j < n; ++j) { scanf("%d",&g[i][j]); } CLR(select,0); ans = INF; for(int i = 0 ; i < n; ++i) DFS(i,1); int tag = 0; for(int i = 0; i < n; ++i) if(ind[i] > 0) if(tag == 0) { printf("%d",i+1); tag = 1; } else printf(" %d",i+1); printf("\n"); } return 0; } int main() { Solve(); return 0; }
{POJ}{3925}{Minimal Ratio Tree}{最小生成树},布布扣,bubuko.com
{POJ}{3925}{Minimal Ratio Tree}{最小生成树}
标签:blog os io strong 文件 for ar div
原文地址:http://www.cnblogs.com/lvpengms/p/3919971.html