hdu 1520Anniversary party(树形dp)

<看完之后感觉要死的算法系列>

Anniversary party


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7539    Accepted Submission(s): 3310




Problem 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
7
1
1
1
1
1
1
1
1 3
2 3
6 4
7 4
4 5
3 5
0 0
 


Sample Output
5

思路：

一些大牛说，图论的题难就难在建图，唉。

用dp[i][0]和dp[i][1]分别表示第i个人不参加与参加。

易得动归方程为：dp[i][1]=dp[j][0];

                                dp[i][0]=max(dp[j][0],dp[j][1]);

其中i和j是上司和下属的关系，若i参加，j不可以参加；若i不参加，j随意。

#include <limits.h>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>#include <algorithm>#include <iostream>#include <iterator>#include <sstream>#include <queue>#include <stack>#include <string>#include <vector>#include <set>#include <map>//#define ONLINE_JUDGE#define eps 1e-6#define INF 0x7fffffff                                          //INT_MAX#define inf 0x3f3f3f3f                                          //int??????????????????#define FOR(i,a) for((i)=0;i<(a);(i)++)                          //[i,a);#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define MEM3(a) memset(a,0x3f,sizeof(a))#define LL __int64const double PI = acos(-1.0);template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }template<class T> inline T Min(T a, T b) { return a < b ? a : b; }template<class T> inline T Max(T a, T b) { return a > b ? a : b; }using namespace std;template<class T>T Mint(T a, T b, T c) {    if (a>b) {        if (c>b)            return b;        return c;    }    if (c>a)        return a;    return c;}template<class T>T Maxt(T a, T b, T c) {    if (a>b) {        if (c>a)            return c;        return a;    }    else if (c > b)        return c;    return b;}const int maxn=6005;int T,n,m;int degree[maxn],val[maxn],head[maxn],vis[maxn],dp[maxn][2];int sum,tot,a,b;//链式前向星方式建图struct node{    int from,to;    int next;}edge[maxn<<1];void addEdge(int a,int b){    edge[tot].from=a,edge[tot].to=b;    edge[tot].next=head[a];    head[a]=tot++;}void dfs(int root){//递归查找该点的贡献度    int ans0,ans1;    ans0=ans1=0;    vis[root]=0;    for(int j=head[root];j!=-1;j=edge[j].next){        int u=edge[j].to;        if(!vis[u])            dfs(u);        ans0+=max(dp[u][0],dp[u][1]);        ans1+=dp[u][0];    }    dp[root][0]=ans0;    dp[root][1]=ans1+val[root];}int main() {    while(~sf(n)){        MEM1(dp);        MEM1(vis);//记录该节点是否已访问        MEM2(head);        MEM1(degree);//记录节点的入度        sum=tot=0;        for2(i,1,n)         sf(val[i]);//记录第i个节点的贡献度        while(sfd(a,b)){            if(!a&&!b) break;            addEdge(b,a);            degree[a]++;        }        for2(i,1,n){            if(!degree[i]){                dfs(i);                sum+=max(dp[i][0],dp[i][1]);            }        }        pf(sum);    }    return 0;}