HDOJ-1561（树形DP）

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一开始感觉这是个森林，怎么DP，突然想到，可以用一个”空“节点作为森林的树根，这个”空“节点没有reward没有cost，其他节点的cost都为1，则状态方程即可写作：f[i][j] = max(f[i][j], f[children[i]][k] + f[i][j-k])

#include <cstdio>#include <vector>#include <algorithm>using namespace std;int N, M;int reward[201] = {0};int gain[201][201] = {0};vector<int> children[201];bool input(){    scanf("%d %d", &N, &M);    if(!N) return false;//input    for(int i = 0; i <= N; ++i) children[i].clear();    for(int i = 1, pa = 0; i <= N; ++i){        scanf("%d %d", &pa, &reward[i]);        children[pa].push_back(i);    }    return true;}void postDP(int i){    int j, k, n = children[i].size() - 1;    for(j = 1; j <= M; ++j) gain[i][j] = reward[i];    for(; n > -1; --n){        int child = children[i][n], cost = !!i;//castle 0 has no cost, others has 1 cost        postDP(child);        for(j = M; j > cost; --j){            for(k = j - cost; k >= 1; --k){                gain[i][j] = max(gain[i][j], gain[child][k] + gain[i][j-k]);            }        }    }}int main(){    while(input()){        if(children[0].empty()){//no castle has no prerequisite            puts("0");            continue;        }        postDP(0);        printf("%d\n", gain[0][M]);    }    return 0;}

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