hdu 1011 树型dp

来源：互联网发布：三轴点胶机编程教学编辑：程序博客网时间：2024/06/05 11:40

#include <cstdio>#include <iostream>#include <cstring>#include <vector>using namespace std;int n,m;int vis[105];int dp[105][105];vector<int> vec[105];vector<int> G[105];int c[105];int p[105];void dfs(int x){vis[x] = 1;int si = G[x].size();for(int i = 0;i < si;i++){int v = G[x][i];if(!vis[v]){dfs(v);vec[x].push_back(v);}}}void  fun(int u){for(int i  = c[u];i <= m;i++){dp[u][i] = p[u];}int si = vec[u].size();for(int i = 0;i < si;i++){int v = vec[u][i];fun(v);for(int j = m;j >= c[u];j --){for(int k = 1;j+k <= m;k++){//这里一定要从1开始因为每个结点必须有一个士兵经过 if(dp[v][k]){dp[u][j+k] = max(dp[u][j+k],dp[v][k]+dp[u][j]);//这里有个隐式的0-1背包问题，dp[i][j]为在第i个结点有jj个士兵经过得到的价值 }}}}}int main(){while(cin >> n  >> m,n!=-1 && m!=-1){for(int i = 1;i <=  n;i++){G[i].clear();vec[i].clear();scanf("%d%d",&c[i],&p[i]);c[i] = (c[i]+19)/20;}for(int i = 0;i < n-1;i++){int x,y;scanf("%d%d",&x,&y);G[x].push_back(y);G[y].push_back(x);}if(m == 0){puts("0");continue;}memset(vis,0,sizeof(vis));dfs(1);memset(dp,0,sizeof(dp));fun(1);cout << dp[1][m] << endl;}return 0;}

0 0