[poj 3342]Party at Hali-Bula 樹形dp

題目鏈接：http://poj.org/problem?id=3342
Party at Hali-Bula

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 5997 Accepted: 2140

Description

Dear Contestant,

I’m going to have a party at my villa at Hali-Bula to celebrate my retirement from BCM. I wish I could invite all my co-workers, but imagine how an employee can enjoy a party when he finds his boss among the guests! So, I decide not to invite both an employee and his/her boss. The organizational hierarchy at BCM is such that nobody has more than one boss, and there is one and only one employee with no boss at all (the Big Boss)! Can I ask you to please write a program to determine the maximum number of guests so that no employee is invited when his/her boss is invited too? I’ve attached the list of employees and the organizational hierarchy of BCM.

Best,
–Brian Bennett

P.S. I would be very grateful if your program can indicate whether the list of people is uniquely determined if I choose to invite the maximum number of guests with that condition.

Input

The input consists of multiple test cases. Each test case is started with a line containing an integer n (1 ≤ n ≤ 200), the number of BCM employees. The next line contains the name of the Big Boss only. Each of the following n-1 lines contains the name of an employee together with the name of his/her boss. All names are strings of at least one and at most 100 letters and are separated by blanks. The last line of each test case contains a single 0.

Output

For each test case, write a single line containing a number indicating the maximum number of guests that can be invited according to the required condition, and a word Yes or No, depending on whether the list of guests is unique in that case.

Sample Input
6
Jason
Jack Jason
Joe Jack
Jill Jason
John Jack
Jim Jill
2
Ming
Cho Ming
0

Sample Output
4 Yes
1 No

Source

Tehran 2006

題意：BB將要邀請BCM的員工參加一個party來慶祝他的退休。還是每個人不能同其直接boss一起參加。要求使參加人數最多，並要求判斷方案是否唯一。

思路：
計算最大的很簡單dp[i][0]=求和max(dp[soni][0],dp[soni][1]); dp[i][1]=求和dp[soni][0];

判斷開始只傻逼的判dp[1][0]=？dp[1][1]；
應用flag[i][0]判斷，如果選的子樹方案不唯一，該方案不唯一;初始當dp[soni][0]==dp[soni][1]時flag【i】[0]=1；

代碼：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<map>
#include<vector>
using namespace std;
int n;
int tot;
vector<int> lin[205];
map<string,int> q;
string s,s1;
int dp[201][2];
int flag[201][2];
void dfs(int x)
{
    int ret=0;
    int tmp=0;
    int f1=0;
    int f2=0;
    for(int i=0;i<lin[x].size();i++)
    {
        int v=lin[x][i];
        dfs(v); 
        ret+=max(dp[v][0],dp[v][1]);

        if(dp[v][0]==dp[v][1]) f1=1;
        else if(dp[v][0]>dp[v][1]&&flag[v][0]==1) f1=1;
        else if(dp[v][1]>dp[v][0]&&flag[v][1]==1) f1=1;

        tmp+=dp[v][0];
        if(flag[v][0]) f2=1;
    }


    dp[x][0]=ret;
    dp[x][1]=tmp+1;
    if(f2) flag[x][1]=1; 
    if(f1) flag[x][0]=1;
}
int main()
{
    while(scanf("%d",&n))
    {
        if(!n) return 0;
        tot=0;
        cin>>s;
        q.clear();
        q[s]=++tot;
        for(int i=1;i<=n;i++)
        {
            lin[i].clear();
        }
        for(int i=1;i<n;i++)
        {
            cin>>s>>s1;
            if(!q[s])
            {
                q[s]=++tot;
            }
            if(!q[s1])
            {
                q[s1]=++tot;
            }
            lin[q[s1]].push_back(q[s]);
        }
        memset(flag,0,sizeof(flag));
        memset(dp,0,sizeof(dp));
        dfs(1);


        if(dp[1][0]==dp[1][1])
        {
            printf("%d No\n",dp[1][0]);
        }
        else if(dp[1][0]>dp[1][1]&&flag[1][0])
        {
            printf("%d No\n",dp[1][0]);
        }
        else if(dp[1][1]>dp[1][0]&&flag[1][1])
        {
            printf("%d No\n",dp[1][1]);
        }
        else printf("%d Yes\n",max(dp[1][0],dp[1][1]));
    }
}