**HDU 4276 - The Ghost Blows Light(树形DP)

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题目：

http://acm.hdu.edu.cn/showproblem.php?pid=4276

题意：

给出一棵树，从1走到n，总时间为T，每走一条边需要花费一定时间，每个结点有一定价值，问在指定时间内回到T的能获取的最大价值。

思路：

先跑一次dfs，得到从1~n必走的路径，将这些路径值置0.

再一次dfs跑树形DP+分组背包，dp[i][j]: 表示到i点花费j时间得到的最大价值。

AC.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 105*2;const int maxt = 505*2;int n, T;int dp[maxn][maxt];int val[maxn];int edge[maxn][maxn], tmpath[maxn];void init(){    memset(edge, -1, sizeof(edge));    memset(tmpath, -1, sizeof(tmpath));    memset(dp, 0, sizeof(dp));}void init_dfs(int u, int fa){    for(int i = 1; i <= n; ++i) {        if(edge[u][i] == -1) continue;        if(i == fa) continue;        tmpath[i] = u;        init_dfs(i, u);    }}void dfs(int u, int fa){    for(int i = 0; i <= T; ++i) dp[u][i] = val[u];    for(int i = 1; i <= n; ++i) {        int w = edge[u][i];        if(w == -1) continue;        if(i == fa) continue;        dfs(i, u);        w = 2*w;        for(int j = T; j >= w; --j) {            for(int k = 0; k <= j - w; ++k) {                dp[u][j] = max(dp[u][j], dp[i][k]+dp[u][j-k-w]);            }        }    }}int main(){    //freopen("in", "r", stdin);    while(~scanf("%d%d", &n, &T)) {        init();        int u, v, w;        for(int i = 1; i < n; ++i) {            scanf("%d%d%d", &u, &v, &w);            edge[u][v] = edge[v][u] = w;        }        for(int i = 1; i <= n; ++i) {            scanf("%d", &val[i]);        }        init_dfs(n, n);        int pre = 1, cnt = 0, sum = val[1];        for(int i = tmpath[1]; ~i; i = tmpath[i]) {            sum += val[i];            cnt += edge[pre][i];            edge[pre][i] = edge[i][pre] = 0;            pre = i;        }        if(cnt > T) {            printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");            continue;        }        T -= cnt;        dfs(1, 1);        printf("%d\n", dp[1][T]);    }    return 0;}

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