poj 2057 The Lost House

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论文题，树dp

多说两句，这个题的决策是在一个节点时选择子节点遍历的顺序

通过贪心来优化dp

#include<cstring>#include<cstdio>#include<vector>#include<cstring>#include<algorithm>using namespace std;const int maxn = 1123;vector<int> edge[maxn];void init(){    for(int i=0;i<maxn;i++){        edge[i].clear();    }}int suc[maxn],fai[maxn],siz[maxn];bool hav[maxn];bool cmper(int x,int y){    return (fai[x] + 2) * siz[y] < (fai[y] + 2) * siz[x];}void dfs(int st){    siz[st] = edge[st].size()==0;    for(vector<int>::iterator it = edge[st].begin();it != edge[st].end();it++){        int x = *it;        dfs(x);        siz[st] += siz[x];    }    sort(edge[st].begin(),edge[st].end(),cmper);    suc[st] = fai[st] = 0;    for(vector<int>::iterator it = edge[st].begin();it != edge[st].end();it++){        int x = *it;        suc[st] += (fai[st]+1)*siz[x] + suc[x];        fai[st] += fai[x] + 2;    }    if(hav[st]) fai[st] = 0;}int main(){    int n;    while(~scanf("%d",&n) && n){        char in[3];        int fa;        init();        for(int i=1;i<=n;i++){            scanf("%d %s",&fa,in);            if(fa != -1) edge[fa].push_back(i);            hav[i] = *in == 'Y';        }        dfs(1);        printf("%.4f\n",suc[1]*1.0/siz[1]);    }    return 0;}

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