HDU 1520Anniversary party 树形dp

A - Anniversary party

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1520

Appoint description: System Crawler  (2016-04-23)

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 T 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 

 dp不难想但是在模型是建立在树上有点麻烦

利用dfs从叶子节点倒着推上去，思维和树塔一样

ACcode：

#include <cstdio>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rd3(x,y,z) scanf("%d%d%d,&x,&y,&z)#define rdl(x) scanf("%I64d,&x)#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int#define ull unsigned long long#define maxn 10005#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)using namespace std;struct tree{    int f,c,b,yes,no;    void init(){        rd(yes);        f=c=b=no=0;    }    int ismax(){        return yes>no?yes:no;    }}my[maxn];void fun(int x){    int dp=my[x].c;    while(dp){        fun(dp);        my[x].yes+=my[dp].no;        my[x].no+=my[dp].ismax();        dp=my[dp].b;    }}int main(){    int n;    while(~rd(n)){        FOR(i,1,n)my[i].init();        int a,b;        while(rd2(a,b),a||b){            my[a].f=b;            my[a].b=my[b].c;            my[b].c=a;        }        FOR(i,1,n)if(!my[i].f){            fun(i);            printf("%d\n",my[i].ismax());            break;        }    }    return 0;}