| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 4995 | Accepted: 2824 |
Description
There is going to be a party to celebrate the 80-th Anniversary of the Ural State University. The University has a hierarchical structure of employees. It means that the supervisor relation forms a tree rooted at the rector V. E. Tretyakov. In order to make the party funny for every one, the rector does not want both an employee and his or her immediate supervisor to be present. The personnel office has evaluated conviviality of each employee, so everyone has some number (rating) attached to him or her. Your task is to make a list of guests with the maximal possible sum of guests' conviviality ratings.
Input
Employees are numbered from 1 to N. A first line of input contains a number N. 1 <= N <= 6 000. Each of the subsequent N lines contains the conviviality rating of the corresponding employee. Conviviality rating is an integer number in a range from -128 to 127. After that go N – 1 lines that describe a supervisor relation tree. Each line of the tree specification has the form: L K It means that the K-th employee is an immediate supervisor of the L-th employee. Input is ended with the line 0 0
Output
Output should contain the maximal sum of guests' ratings.
Sample Input
711111111 32 36 47 44 53 50 0
Sample Output
5 树形DP： 先建图，注意后面的点才是父亲，根不一定是1。 设F[x][1]表示第x个人来聚会的最大欢乐值，F[x][0]表示他不来的欢乐值，所以就可以写出转移方程: F[x][1] += F[son(x)][0]; F[x][0] += max(F[som(x)][1], F[son(x)][0]); CODE:
#include#include #include #define REP(i, s, n) for(int i = s; i <= n; i ++)#define REP_(i, s, n) for(int i = n; i >= s; i --)#define max(a, b) a > b ? a : b#define MAX_N 6000 + 10using namespace std;int n, u, v, R;int F[MAX_N][2];struct node{ int v, next;}E[MAX_N << 1];int head[MAX_N], top = 0;void add(int u, int v){ E[++ top].v = v; E[top].next = head[u]; head[u] = top;}void Gao(int x){ for(int i = head[x]; i; i = E[i].next){ Gao(E[i].v); F[x][1] += F[E[i].v][0]; F[x][0] += max(F[E[i].v][1], F[E[i].v][0]); }}int main(){ scanf("%d", &n); REP(i, 1, n) scanf("%d", &F[i][1]); bool ok[MAX_N]; memset(ok, 0, sizeof(ok)); while(scanf("%d%d", &u, &v) != EOF){ if(u == 0 && v == 0) break; ok[u] = 1; add(v, u); } REP(i, 1, n) if(!ok[i]) { R = i; break; } Gao(R); printf("%d\n", max(F[R][1], F[R][0])); return 0;}