Jury Compromise

K - Jury Compromise

参考：ACM POJ 1015 Jury Compromise（陪审团的人选，动态规划题，难）

说实话真有点难想，用一个DP[i][j]来表示在选取i个人，辩控差为j（j值已做些许处理）时辩控总分的最大值，用三个for循环来更新这个值。具体思路还是看参考博客吧....

优先队列默认top()是最大值，如果写成priority_queue<int,vector<int>,greater<int> > top()则为最小值

代码：

// Created by CAD on 2019/10/30.
#include <iostream>
#include <cstring>
#include <queue>
#include <vector>
#define mst(name, value) memset(name,value,sizeof(name))
using namespace std;

int dp[25][805],pre[25][805];
struct score{
    int a,b;
}p[205];
priority_queue<int,vector<int>,greater<int> > q;
void print(int m,int n)
{
    if(!pre[m][n]) return ;
    print(m-1,n-(p[pre[m][n]].a-p[pre[m][n]].b));
    q.push(pre[m][n]);
}
int main()
{
//    FOPEN;
    ios::sync_with_stdio(false);
    cin.tie(0);
    int m,n;
    int Case=0;
    while(cin>>n>>m,m|n)
    {
        for(int i=1;i<=n;++i)
            cin>>p[i].a>>p[i].b;
        mst(dp,-1);
        mst(pre,0);
        int full=m*20;
        dp[0][full]=0;
        for(int i=0;i<m;++i)
            for(int j=0;j<=full*2;++j)
                if(dp[i][j]>=0)
                    for(int k=1;k<=n;++k)
                    {
                        if(dp[i][j]+p[k].a+p[k].b>dp[i+1][j+p[k].a-p[k].b])
                        {
                            int a=i,b=j;
                            while(a>0&&pre[a][b]!=k)
                                b-=(p[pre[a][b]].a-p[pre[a][b]].b),a--;
                            if(!a)
                                dp[i+1][j+p[k].a-p[k].b]=dp[i][j]+p[k].a+p[k].b,
                                pre[i+1][j+p[k].a-p[k].b]=k;
                        }
                    }
        int temp=0;
        while(dp[m][full+temp]<0&&dp[m][full-temp]<0) temp++;
        if(dp[m][full+temp]>dp[m][full-temp])   temp+=full;
        else temp=full-temp;
        cout<<"Jury #"<<++Case<<endl;
        cout<<"Best jury has value "<< (temp-full+dp[m][temp])/2 <<
        " for prosecution and value "<<(dp[m][temp]-temp+full)/2<<" for defence: "<<endl;
        print(m,temp);
        while(!q.empty())
            cout<<" "<<q.top(),q.pop();
        cout<<endl<<endl;
    }
    return 0;
}