poj1463树形dp

来源：互联网发布：管家婆远程数据库编辑：程序博客网时间：2024/06/05 04:20

题目大意：最少需要多少个点上方士兵，才能观察到所有边。

思路：dp[i][0]表示节点上不放士兵，dp[i][1] 表示节点i放士兵。那么dp[i]均可由子树得到。dp[i][0] = dp[j0][1]+dp[j1][1] + …… , dp[i][1] = min(dp[j0][1] , dp[j0][0] ) + min(dp[j1][0] , dp[j1][1]) + ……因为当前节点放士兵时子节点可放可不放（很重要，刚开始疏忽了）。

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <fstream>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#include <iomanip>using namespace std;//#pragma comment(linker, "/STACK:102400000,102400000")#define maxn 1505#define MOD 1000000007#define mem(a , b) memset(a , b , sizeof(a))#define LL long long#define INF 100000000int n , m;int vis[maxn];vector<int>v[maxn];char str[10];int dp[maxn][2];void dfs(int root){    int up = v[root].size();    if(up == 0) {dp[root][0] = 0 , dp[root][1] = 1; return;}    vis[root] = 0;    for(int i = 0 ; i < up ; i ++)    {        if(!vis[v[root][i]]) dfs(v[root][i]);        dp[root][0] += dp[v[root][i]][1];        dp[root][1] += min(dp[v[root][i]][0] , dp[v[root][i]][1]);    }}int main(){    while(scanf("%d" , &n) != EOF)    {        mem(dp , 0);mem(vis , 0);        for(int i = 0 ;  i <= n ; i ++) v[i].clear() , dp[i][1] = 1;        for(int i = 0 ; i < n ; i ++)        {            scanf("%s" , str);            int up = 0 , id = 0 , e , tmp = 0;            while(str[id] != ':') tmp = tmp * 10 + (str[id++] - '0') ;            id ++;id++;            while(str[id] != ')') up = up * 10 + (str[id++] - '0');           // cout << up << " " << tmp << endl;            for(int j = 0 ; j < up ; j ++)            {                int u = tmp;                scanf("%d" , &e);                if(u > e) swap(u , e);                v[u].push_back(e);             //   cout << e << endl;            }        }        //cout << "test" << endl;        dfs(0);        printf("%d\n" , min(dp[0][0] , dp[0][1]));    }}

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