1015--Jury Compromise

题意：从n个人中选出m个，每个人有固定的p值，d值，要求使m个人的p总和和d总和的差的绝对值最小，若有多解则取两者和最大的。

此题本质上可以看成是背包问题，差和的绝对值最小即是差和小于等于0的值最大，大于等于0的值最小。对于前者，可以看成是背包容量为0时可获得的最大价值；对于后者，交换差值的减数和被减数，即可化为前一种情况。题目还要求出路径，因此，还需要同时记录最大的和。

#include <iostream>using namespace std;int m,n;int D[210],P[210];int a[210];int t[201][21][401]={0};int z[201][21][401]={0};int G[210],H[210];int MAX(int x,int y){return x>y?x:y;}int main(){cin>>n>>m;int idx=0;while(n>0){idx++;int vmax=-400;for(int i=1;i<=n;i++){cin>>D[i]>>P[i];a[i]=D[i]-P[i]+20;}for(int i=0;i<=n;i++)for(int j=0;j<=m;j++)for(int v=0;v<=20*m;v++){t[i][j][v]=-400; z[i][j][v]=-400;if(i==0  ){t[i][j][v]=-400;z[i][j][v]=-400;}if(j>i) {t[i][j][v]=-400;z[i][j][v]=-400;}if(j==0 ){t[i][0][0]=0;z[i][0][0]=0;}t[0][0][0]=0; z[0][0][0]=0;}for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)for(int v=0;v<=20*m;v++){if(v-a[i]>=0 && v-a[i]<=20*m){t[i][j][v]=MAX(t[i-1][j][v],t[i-1][j-1][v-a[i]]+D[i]+P[i]);z[i][j][v]=MAX(z[i-1][j][v],z[i-1][j-1][v-a[i]]+a[i]);if(j==m && z[i][j][v]>vmax){vmax=z[i][j][v];}}else{if(t[i][j][v]<t[i-1][j][v] && t[i-1][j][v]>0)t[i][j][v]=t[i-1][j][v];if(z[i][j][v]<z[i-1][j][v] && z[i-1][j][v]>0)z[i][j][v]=z[i-1][j][v];if(j==m && z[i][j][v]>vmax){vmax=z[i][j][v];}}}for(int i=0;i<=n;i++){G[i]=0;H[i]=0;}int dpmax=0,d=0,p=0,dp=0;int ii=n,v=vmax,j=m;int tmax=0,tidx=0;while(j>0 && ii>0){if(t[ii][j][v]==t[ii-1][j][v]){G[ii]=0;}else {G[ii]=1;d+=D[ii];p+=P[ii];dp+=D[ii]+P[ii];v=v-a[ii];j--;}ii--;}int vmax2=-400;for(int i=1;i<=n;i++){a[i]=P[i]-D[i]+20;}for(int i=0;i<=n;i++)for(int j=0;j<=m;j++)for(int v=0;v<=20*m;v++){t[i][j][v]=-400; z[i][j][v]=-400;if(i==0  ){t[i][j][v]=-400;z[i][j][v]=-400;}if(j>i) {t[i][j][v]=-400;z[i][j][v]=-400;}if(j==0 ){t[i][0][0]=0;z[i][0][0]=0;}t[0][0][0]=0; z[0][0][0]=0;}for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)for(int v=0;v<=20*m;v++){if(v-a[i]>=0 && v-a[i]<=20*m){t[i][j][v]=MAX(t[i-1][j][v],t[i-1][j-1][v-a[i]]+D[i]+P[i]);z[i][j][v]=MAX(z[i-1][j][v],z[i-1][j-1][v-a[i]]+a[i]);if(j==m && z[i][j][v]>vmax2){vmax2=z[i][j][v];}}else{if(t[i][j][v]<t[i-1][j][v] && t[i-1][j][v]>0)t[i][j][v]=t[i-1][j][v];if(z[i][j][v]<z[i-1][j][v] && z[i-1][j][v]>0)z[i][j][v]=z[i-1][j][v];if(j==m && z[i][j][v]>vmax){vmax=z[i][j][v];}}}int dpmax2=0,d2=0,p2=0,dp2=0;ii=n;v=vmax2;j=m;while(j>0 && ii>0){if(t[ii][j][v]==t[ii-1][j][v]){H[ii]=0;}else {H[ii]=1;d2+=D[ii];p2+=P[ii];dp2+=D[ii]+P[ii];v=v-a[ii];j--;}ii--;}cout<<"Jury #"<<idx<<endl;if(vmax>vmax2 || (vmax==vmax2 && dp>dp2)){cout<<"Best jury has value "<<d<<" for prosecution and value "<<p<<" for defence:"<<endl; for(int i=1;i<=n;i++)if(G[i]) cout<<" "<<i;cout<<endl<<endl;}else{cout<<"Best jury has value "<<d2<<" for prosecution and value "<<p2<<" for defence:"<<endl; for(int i=1;i<=n;i++)if(H[i]) cout<<" "<<i;cout<<endl<<endl;}cin>>n>>m;}return 0;}