poj 3342 Party at Hali-Bula（树形DP+判断方式是不是唯一）

1、http://poj.org/problem?id=3342

2、题目大意：

现在要邀请n个人中的一些人参加晚宴，要求是有直接上下级关系的人不能同时出席，问最多可以邀请多少人参加，并判断在保证最大人数的情况下，人是不是唯一确定的

状态转移方程很好确定，难在怎么判断方案是不是唯一，假设第i个人参加时值最大，那么不能确定是不是不唯一，但是如果第i个人不去的方案最优的话，他的状态有两种，孩子要么去，要么不去，如果这两种状态的最大值相同，那么就是不唯一的，详见代码

3、题目：

Party at Hali-Bula

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 4917 Accepted: 1741

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 integern (1 ≤n ≤ 200), the number of BCM employees. The next line contains the name of the Big Boss only. Each of the followingn-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

6JasonJack JasonJoe JackJill JasonJohn JackJim Jill2MingCho Ming0

Sample Output

4 Yes1 No

 

4、AC代码：

#include<stdio.h>#include<string.h>#include<string>#include<algorithm>#include<map>#include<vector>using namespace std;vector<int> vec[205];map<string,int> mp;char str[205][105];char tmp[105];char a[105],b[105];int dp[205][2];int k;int find(char a[105]){    for(int i=0; i<k; i++)    {        if(strcmp(a,str[i])==0)            return i;    }    strcpy(str[k++],a);    return k-1;}void dfs(int root){    for(int i=0; i<vec[root].size(); i++)    {        int v=vec[root][i];        dfs(v);        dp[root][0]+=max(dp[v][0],dp[v][1]);        dp[root][1]+=dp[v][0];    }}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        if(n==0)            break;        k=0;        mp.clear();        scanf("%s",tmp);        mp[tmp]=k++;        for(int i=0; i<n; i++)        {            vec[i].clear();            dp[i][0]=0;            dp[i][1]=1;        }        for(int i=1; i<n; i++)        {            scanf("%s%s",a,b);            if(mp.find(a)==mp.end()) mp[a]=k++;            if(mp.find(b)==mp.end()) mp[b]=k++;            vec[mp[b]].push_back(mp[a]);        }        dfs(0);        //下面代码判断是不是唯一的方案        int flag=1;        for(int i=0; i<n; i++)        {            if(dp[i][0]>dp[i][1])            {                for(int j=0; j<vec[i].size(); j++)                {                    int v=vec[i][j];                    if(dp[v][0]==dp[v][1])                    {                        flag=0;                        break;                    }                }            }            if(flag==0)            break;        }        if(flag==0 || dp[0][0]==dp[0][1])//注意判断条件，            printf("%d No\n",max(dp[0][0],dp[0][1]));        else            printf("%d Yes\n",max(dp[0][0],dp[0][1]));    }    return 0;}/*6JasonJack JasonJoe JackJill JasonJohn JackJim Jill2MingCho Ming0*/

5、AC代码,处理字符串用的for循环遍历的，没有用map

 

#include<stdio.h>#include<string.h>#include<vector>using namespace std;vector<int> vec[205];char str[205][105];char tmp[105];char tmp1[105];int dp[205][2];int k;int find(char a[105]){    for(int i=0; i<k; i++)    {        if(strcmp(a,str[i])==0)            return i;    }    strcpy(str[k++],a);    return k-1;}void dfs(int root){    for(int i=0; i<vec[root].size(); i++)    {        int v=vec[root][i];        dfs(v);        dp[root][0]+=max(dp[v][0],dp[v][1]);        dp[root][1]+=dp[v][0];    }}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        if(n==0)            break;        k=0;        scanf("%s",tmp);        strcpy(str[k++],tmp);        for(int i=0; i<n; i++)        {            vec[i].clear();            dp[i][0]=0;            dp[i][1]=1;        }        for(int i=1; i<n; i++)        {            scanf("%s%s",tmp,tmp1);            int a=find(tmp);            int b=find(tmp1);            vec[b].push_back(a);        }        dfs(0);        int flag=0;        for(int i=0; i<n; i++)        {            if(dp[i][0]>=dp[i][1])            {                for(int j=0; j<vec[i].size(); j++)                {                    int v=vec[i][j];                    if(dp[v][0]==dp[v][1])                    {                        flag=1;                        break;                    }                }            }            if(flag==1)            break;        }        if(flag==1 || dp[0][0]==dp[0][1])            printf("%d No\n",max(dp[0][0],dp[0][1]));        else            printf("%d Yes\n",max(dp[0][0],dp[0][1]));    }    return 0;}/*6JasonJack JasonJoe JackJill JasonJohn JackJim Jill2MingCho Ming0*/