zoj3715 Kindergarten Election

题意： 有n（3-100）个人 ，给定n-1个人的投票以及动摇他们投票给自己的代价，问，如何使用最少的代价赢得选举

解题思路：本来开始是想如何一步步贪心得到最优解，后来才知道这属于多变量问题，解法就是枚举其中某些变量的值然后求其他变量的最优解来找出最优解

解题代码：

// File Name: k.cpp
// Author: darkdream
// Created Time: 2013年05月18日 星期六 14时31分10秒

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>

using namespace std;
int piao[105],cost[105],people[105];
struct node
{
    int num;
    int site;
    bool operator < (const node &x) const 
    {
        return num < x.num;
    }
}nodes[105];
int main()
{

    int t ;
//    freopen("/home/plac/problem/input.txt","r",stdin);
    scanf("%d",&t);
    while(t--)
    {
        memset(piao,0,sizeof(piao));
        memset(cost,0,sizeof(cost));
        memset(people,0,sizeof(people));
        int n ,Max = 10000000;
        scanf("%d",&n);
        for(int i= 2;i <=n; i++)
        {
            scanf("%d",&people[i]);
            piao[people[i]]++;
        }
        for(int i = 2; i <=n; i ++)
            scanf("%d",&cost[i]);
        int people1[105];
        for(int may = (piao[1] == 0 || piao[1] == 1) ? 2:piao[1] ; may < n; may ++)
        {
            memset(nodes,0,sizeof(nodes));
            int sum = 0;
            int temp = piao[1];
            for(int i = 2; i <= n; i++)
                if(piao[i] > may -1)
                    sum+= piao[i]-(may-1);
            if(sum + piao[1] > may)
                continue;
            for(int i = 0 ; i <= n;i ++)
                people1[i] = people[i];
            int j = 0 , Sum = 0 ;
            for(int i = 2; i <=n; i++)
            {
                if(piao[i] > may-1)
                { j = 0;
                    for( int k = 2; k <= n; k ++)
                        if(people1[k] == i)
                        {
                            nodes[j].num = cost[k];
                            nodes[j].site = k ;
                            j++;
                        }
                   sort(nodes,nodes+j);
                    for(int k = 0; k < piao[i]-(may-1); k++  )
                     {
                    Sum += nodes[k].num;
                    people1[nodes[k].site] = 1;
                     piao[1]++;
                   }
 
                }

            }
 


            memset(nodes,0,sizeof(nodes));
            j = 0;
            for(int  i = 2 ;i <= n;i ++)
            {
               if(people1[i] != 1)
               {
                   nodes[j].num = cost[i];
                   nodes[j].site  = i ;
                   j++;
               }

            }

            sort(nodes,nodes+j);
            for(int k = 0; k < j ; k++  )
            {
                if(piao[1] >= may)
                    break;
                Sum += nodes[k].num;
                people1[nodes[k].site] = 1;
                piao[1]++;
            }

            if(Sum < Max)
            {    
                Max = Sum;
            
            }
            piao[1] = temp;
        }
    printf("%d\n",Max);

    }
    return  0 ;

}