POJ-1015(背包变形+输出路径)

Jury Compromise

POJ-1015

• 推荐几个较好的介绍和理解：https://blog.csdn.net/lyy289065406/article/details/6671105
• https://blog.csdn.net/glqac/article/details/22687243

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
const int INF=0X3F3F3F3F;
const int maxn=203;
const int maxm=22;
int n,m;
int d[maxn],p[maxn];//被告和原告
int add[maxn],sub[maxn];//sub[i]表示辩控差也就是p-d
int dp[maxn][maxm][802];//dp[i][j][k]表示从前i个人里面选择j个人，辩控差为k时的最大值// dp[i,j,k]=max{dp[i-1,j,k],dp[i-1,j-1,k-d[i]]+ad[i]}
int path[maxn][maxm][802];
int ans[maxm];
struct node{
    int defence;
    int prosecute;
    int num;
};
int main(){
    int cases=0;
    while(cin>>n>>m&&(n||m)){
        for(int i=1;i<=n;i++){
            cin>>p[i]>>d[i];
            add[i]=p[i]+d[i];
            sub[i]=p[i]-d[i];
        }
        memset(dp,-INF,sizeof(dp));
        int fix=m*20;//表示修正值
        for(int i=0;i<=n;i++){
            dp[i][0][fix]=0;////由于修正了数值，因此dp[i][0][fix]才是真正的dp[i][0][0]
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=i&&j<=m;j++){
                for(int k=0;k<=fix*2;k++){
                    dp[i][j][k]=dp[i-1][j][k];
                    path[i][j][k]=path[i-1][j][k];
                    if(k>=sub[i]&&dp[i-1][j-1][k-sub[i]]>=0&&k-sub[i]<=2*fix){
                        if(dp[i-1][j-1][k-sub[i]]+add[i]>dp[i][j][k]){
                            dp[i][j][k]=dp[i-1][j-1][k-sub[i]]+add[i];
                            path[i][j][k]=i;
                        }
                    }
                }
            }
        }
        int diff,k;
        for(k=0;k<=fix;k++){
            if(dp[n][m][fix-k]>=0||dp[n][m][fix+k]>=0){
                break;
            }
        }
        if(dp[n][m][fix-k]>dp[n][m][fix+k]){
            diff=fix-k;
        }else diff=fix+k;
        cout<<"Jury #"<< ++cases <<endl;
        cout<< "Best jury has value ";
        //辩方总值 = （辨控和+辨控差-修正值）/2
		cout<<(dp[n][m][diff]+diff-fix)/2<<" for prosecution and value ";

        //控方总值 = （辨控和-辨控差+修正值）/2
		cout<<(dp[n][m][diff]-diff+fix)/2<<" for defence:"<<endl;
        //类似于背包输出路径
        for (int i=n,j=m,k=diff; j>=1;)
		{
			int p = path[i][j][k] ;
			ans[j] = p;
			k -= sub[p];
			j--;
			i = path[p-1][j][k];//
		}
		for (int i=1; i<=m; i++)
		{
			cout << " " << ans[i];
		}
        cout<<endl<<endl;
    }
    return 0;
}