POJ-1947 Rebuilding Roads (树上背包)

Rebuilding Roads

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 12245 Accepted: 5673

Description

The cows have reconstructed Farmer John's farm, with its N barns (1 <= N <= 150, number 1..N) after the terrible earthquake last May. The cows didn't have time to rebuild any extra roads, so now there is exactly one way to get from any given barn to any other barn. Thus, the farm transportation system can be represented as a tree. 

Farmer John wants to know how much damage another earthquake could do. He wants to know the minimum number of roads whose destruction would isolate a subtree of exactly P (1 <= P <= N) barns from the rest of the barns.

Input

* Line 1: Two integers, N and P 

* Lines 2..N: N-1 lines, each with two integers I and J. Node I is node J's parent in the tree of roads. 

Output

A single line containing the integer that is the minimum number of roads that need to be destroyed for a subtree of P nodes to be isolated. 

Sample Input

11 61 21 31 41 52 62 72 84 94 104 11

Sample Output

2

#include <stdio.h>#include <vector>#include <string.h>using namespace std;vector<vector<int> > g(151);int dp[151][151], pre[151], n, p;void dfs(int x){for(int i = 0; i <= p; ++i){dp[x][i] = 1e9;}dp[x][1] = 0;  //叶子节点，不包括父亲节点，不需要切就可得1个int cur;for(int i = 0; i < g[x].size(); ++i){cur = g[x][i];dfs(cur);for(int j = p; j >= 0; --j){ dp[x][j] += 1;   //不包括当前儿子for(int s = 1; s < j; ++s){  //包括当前儿子dp[x][j] = min(dp[x][j], dp[x][j - s] + dp[cur][s]);}}}}int main(){int u, v;scanf("%d %d", &n, &p);memset(pre, 0, sizeof(pre));for(int i = 1; i < n; ++i){scanf("%d %d", &u, &v);g[u].push_back(v);pre[v] = u;}int rt;for(int i = 1; i <= n; ++i){if(pre[i] == 0){rt = i;break;}}dfs(rt);int ans = dp[rt][p];for(int i = 1; i <= n; ++i){ans = min(ans, dp[i][p] + 1);}printf("%d\n", ans);}/*题意：一棵树，150个节点，现在想得到一棵有p个节点的子树，问最少需要切多少边才可以得到。思路：树上背包，dp[i][j]表示以i为节点的子树得到节点数为j的子树最少需要切多少刀。这里不包括父亲节点的部分。对于每个节点对所有儿子节点跑一下背包即可。*/