HDU 5834 Magic boy Bi Luo with his excited tree 树形DP ★ ★

转载于：http://www.cnblogs.com/qscqesze/p/5771193.html

题意

给你一棵树，边有边权，每经过边一次，就得支付过路费c[i]，点上面有宝藏，每个点只能拿一次。

问从每个点出发，能够拿到的最大值是多少？

题解：

显然的树形DP

dfs两次，第一次dfs维护从这个点的子树出去，再回来能够取得的最大值是多少，不回来的最大值是多少。

然后第二次dfs，再加上从fa那儿转移过来的值就好了，同样维护回来，和不回来。

显然答案就是max(从fa回来+从子树不回来,从fa不回来+从子树回来）

树形DP维护一下这些玩意儿就好了。

#include <bits/stdc++.h>#define INF 1000000007#define FI first#define SE second#define PB emplace_back#define VI vector<int>using namespace std;typedef long long LL;typedef pair <int, int> P;const int NUM = 100010;struct edge {int next, to, cost;} e[NUM * 2];int head[NUM], tot;void gInit() {memset(head, -1, sizeof(head)); tot = 0;}void add_edge(int u, int v, int c){    e[tot] = (edge) {head[u], v, c};    head[u] = tot++;}int n, V[NUM], w1[NUM], w2[NUM], id[NUM], ans[NUM];//w1是从该点出发最后回到该点的最大值//w2是从该点出发有一次不回来的最大值//id是不回来的路的子树根节点void dfs1(int u, int fa){    w1[u] = V[u];    w2[u] = V[u];    id[u] = -1;    for(int i = head[u]; ~i; i = e[i].next) {        int v = e[i].to;        if(v == fa) continue;        dfs1(v, u);        int tmp = max(0, w1[v] - e[i].cost - e[i].cost);        int tmmp = w1[u] + max(0, w2[v] - e[i].cost);        w2[u] += tmp;        if(w2[u] < tmmp) {            w2[u] = tmmp;            id[u] = v;        }        w1[u] += tmp;    }}//sum1同w1,sum2同w2void dfs2(int u, int fa, int sum1, int sum2){    ans[u] = max(w1[u] + sum2, w2[u] + sum1);    int W1 = w1[u];    int W2 = w2[u];    int Id = id[u];    W2 += sum1;    if(W2 <= W1 + sum2) {        W2 = W1 + sum2;        Id = fa;    }    W1 += sum1;    for(int i = head[u]; ~i; i = e[i].next) {        int v = e[i].to;        if(v == fa) continue;        if(v == Id) {            int tmp1 = sum1 + V[u], tmp2 = sum2 + V[u];            for(int j = head[u]; ~j; j = e[j].next) {                int vv = e[j].to;                if(vv == fa || v == vv) continue;                int tmp = max(0, w1[vv] - e[j].cost - e[j].cost);                tmp2 = max(tmp1 + max(0, w2[vv] - e[j].cost), tmp2 + tmp);                tmp1 += tmp;            }            tmp1 = max(0, tmp1 - e[i].cost - e[i].cost);            tmp2 = max(0, tmp2 - e[i].cost);            dfs2(v, u, tmp1, tmp2);        }        else {            int tmp = max(0, w1[v] - e[i].cost - e[i].cost);            int tmp1 = max(0, W1 - tmp - e[i].cost - e[i].cost);            int tmp2 = max(0, W2 - tmp - e[i].cost);            dfs2(v, u, tmp1, tmp2);        }    }}int main(){    int T; scanf("%d", &T);    for(int cas = 1; cas <= T; ++cas) {        gInit();        scanf("%d", &n);        for(int i = 1; i <= n; ++i) {            scanf("%d", &V[i]);        }        for(int i = 1; i < n; ++i) {            int u, v, c;            scanf("%d%d%d", &u, &v, &c);            add_edge(u, v, c);            add_edge(v, u, c);        }        dfs1(1, -1);        dfs2(1, -1, 0, 0);        printf("Case #%d:\n", cas);        for(int i = 1; i <= n; ++i)            printf("%d\n", ans[i]);    }    return 0;}