首页 > 代码库 > HDU2489 Minimal Ratio Tree 【DFS】+【最小生成树Prim】
HDU2489 Minimal Ratio Tree 【DFS】+【最小生成树Prim】
Minimal Ratio Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2382 Accepted Submission(s): 709
Problem Description
For a tree, which nodes and edges are all weighted, the ratio of it is calculated according to the following equation.
Given a complete graph of n nodes with all nodes and edges weighted, your task is to find a tree, which is a sub-graph of the original graph, with m nodes and whose ratio is the smallest among all the trees of m nodes in the graph.
Given a complete graph of n nodes with all nodes and edges weighted, your task is to find a tree, which is a sub-graph of the original graph, with m nodes and whose ratio is the smallest among all the trees of m nodes in the graph.
Input
Input contains multiple test cases. The first line of each test case contains two integers n (2<=n<=15) and m (2<=m<=n), which stands for the number of nodes in the graph and the number of nodes in the minimal ratio tree. Two zeros end the input. The next line contains n numbers which stand for the weight of each node. The following n lines contain a diagonally symmetrical n×n connectivity matrix with each element shows the weight of the edge connecting one node with another. Of course, the diagonal will be all 0, since there is no edge connecting a node with itself.
All the weights of both nodes and edges (except for the ones on the diagonal of the matrix) are integers and in the range of [1, 100].
The figure below illustrates the first test case in sample input. Node 1 and Node 3 form the minimal ratio tree.
All the weights of both nodes and edges (except for the ones on the diagonal of the matrix) are integers and in the range of [1, 100].
The figure below illustrates the first test case in sample input. Node 1 and Node 3 form the minimal ratio tree.
Output
For each test case output one line contains a sequence of the m nodes which constructs the minimal ratio tree. Nodes should be arranged in ascending order. If there are several such sequences, pick the one which has the smallest node number; if there‘s a tie, look at the second smallest node number, etc. Please note that the nodes are numbered from 1 .
Sample Input
3 2 30 20 10 0 6 2 6 0 3 2 3 0 2 2 1 1 0 2 2 0 0 0
Sample Output
1 3 1 2
Source
2008 Asia Regional Beijing
⊙﹏⊙‖∣在判断两浮点数大小时应该这样比较:a-b<-(1e-8);我因为不知道这个WA了6次。题意:求一个略微变形的“最小生成树”,其值为边权和除以点权和。
题解:用深搜在n个点里选出m个点,再求这m个点的“最小生成树”即可。
#include <stdio.h> #include <string.h> #include <limits.h> #define maxn 16 int map[maxn][maxn], node[maxn]; int n, m, store[maxn], vis[maxn]; double ans; bool visted[maxn]; //hash to vis array double prim() { int i, j, u, count = 0, tmp, vnv = 0, vne = 0; for(i = 1; i <= m; ++i) vnv += node[vis[i]]; memset(visted, 0, sizeof(visted)); visted[1] = 1; while(count < m - 1){ for(i = 1, tmp = INT_MAX; i <= m; ++i){ if(!visted[i]) continue; for(j = 1; j <= m; ++j){ if(!visted[j] && map[vis[i]][vis[j]] < tmp){ tmp = map[vis[i]][vis[j]]; u = j; } } } if(tmp != INT_MAX){ visted[u] = 1; vne += tmp; ++count; } } return vne * 1.0 / vnv; } void DFS(int k, int id) { if(id > m){ double tmp = prim(); if(tmp - ans < -(1e-8)){ ans = tmp; memcpy(store, vis, sizeof(vis)); } return; } for(int i = k; i <= n; ++i){ vis[id] = i; DFS(i + 1, id + 1); } } int main() { int i, j; while(scanf("%d%d", &n, &m), n || m){ for(i = 1; i <= n; ++i) scanf("%d", &node[i]); for(i = 1; i <= n; ++i) for(j = 1; j <= n; ++j) scanf("%d", &map[i][j]); ans = INT_MAX; DFS(1, 1); for(i = 1; i <= m; ++i) if(i != m) printf("%d ", store[i]); else printf("%d\n", store[i]); } return 0; }
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