标签:hdu2489 minimal ratio tree 最小生成树 prim dfs
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
#include <cstdio> #include <cstring> #include <algorithm> #define MAX 20 #define INF 100000000 using namespace std ; int graph[MAX][MAX] , lowCost[MAX] , node[MAX] , path[MAX] , temp[MAX]; int n ,m ; bool cmp(const int a , const int b) { return a<b ; } int prim(int s) { bool visited[MAX] ; memset(visited,false,sizeof(visited)) ; int sum = 0 ; for(int i = 0 ; i < m ; ++i) { lowCost[temp[i]] = graph[s][temp[i]] ; } visited[s] = true ; for(int i = 0 ; i < m-1 ; ++i) { int min = INF , index = -1 ; for(int j = 0 ; j < m ; ++j) { if(!visited[temp[j]] && lowCost[temp[j]]<min) { min = lowCost[temp[j]] ; index = temp[j] ; } } if(index == -1) { break ; } visited[index] = true ; sum += min ; for(int j = 0 ; j < m ; ++j) { if(!visited[temp[j]] && lowCost[temp[j]]>graph[index][temp[j]]) { lowCost[temp[j]] = graph[index][temp[j]] ; } } } return sum ; } double ans = INF*1.0 ; void DFS(int num , int count) { if(count == m) { int sumOfEdge = prim(temp[0]) ,sumOfNode = 0; for(int i = 0 ; i < m ; ++i) { sumOfNode += node[temp[i]] ; } double t = sumOfEdge*1.0/sumOfNode ; if(ans-t>0.00001) { ans = t ; for(int i = 0 ; i < m ; ++i) { path[i] = temp[i] ; } } return ; } for(int i = num+1 ; i <= n ; ++i) { temp[count] = i ; DFS(i,count+1); } } int main() { while(~scanf("%d%d",&n,&m) && (m||n)) { 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",&graph[i][j]) ; } graph[i][i] = INF ; } ans = INF*1.0 ; for(int i = 1 ; i <= n ; ++i) { temp[0] = i ; DFS(i,1); } sort(path,path+m,cmp) ; for(int i = 0 ; i < m ; ++i) { printf("%d",path[i]) ; if(i != m-1) { printf(" ") ; } } puts("") ; } return 0 ; }
hdu 2489 Minimal Ratio Tree DFS枚举点+最小生成树 属于中等偏上题 ,Double比较大小的时候注意精度问题
标签:hdu2489 minimal ratio tree 最小生成树 prim dfs
原文地址:http://blog.csdn.net/lionel_d/article/details/43919601