POJ 1015 Jury Compromise(双塔dp)
Time Limit: 1000MS | Memory Limit: 65536K | |||
Total Submissions: 33737 | Accepted: 9109 | Special Judge |
Description
Based on the grades of the two parties, the judge selects the jury. In order to ensure a fair trial, the tendencies of the jury to favour either defence or prosecution should be as balanced as possible. The jury therefore has to be chosen in a way that is satisfactory to both parties.
We will now make this more precise: given a pool of n potential jurors and two values di (the defence's value) and pi (the prosecution's value) for each potential juror i, you are to select a jury of m persons. If J is a subset of {1,..., n} with m elements, then D(J ) = sum(dk) k belong to J
and P(J) = sum(pk) k belong to J are the total values of this jury for defence and prosecution.
For an optimal jury J , the value |D(J) - P(J)| must be minimal. If there are several jurys with minimal |D(J) - P(J)|, one which maximizes D(J) + P(J) should be selected since the jury should be as ideal as possible for both parties.
You are to write a program that implements this jury selection process and chooses an optimal jury given a set of candidates.
Input
These values will satisfy 1<=n<=200, 1<=m<=20 and of course m<=n. The following n lines contain the two integers pi and di for i = 1,...,n. A blank line separates each round from the next.
The file ends with a round that has n = m = 0.
Output
On the next line print the values D(J ) and P (J ) of your jury as shown below and on another line print the numbers of the m chosen candidates in ascending order. Output a blank before each individual candidate number.
Output an empty line after each test case.
Sample Input
4 2
1 2
2 3
4 1
6 2
0 0
Sample Output
Jury #1
Best jury has value 6 for prosecution and value 4 for defence:
2 3
Hint
Source
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = ;
const int inf = 2.1e9;
const ll Inf = ;
const int mod = ;
const double eps = 1e-;
const double pi = acos(-);
int n,m;
int dp[][];
int num1[],num2[],sum[],diff[];
int path[][];
int fix = ; bool check(int i,int k,int j){
while(i&&path[i][k]!=j){
k-=diff[path[i][k]];
i--;
}
return i==;
} int main()
{
// ios::sync_with_stdio(false);
// freopen("in.txt","r",stdin);
int cases=;
while(scanf("%d%d",&n,&m)!=EOF&&(n||m)){
vector<int>ans;
for(int i=;i<=n;i++){
scanf("%d%d",&num1[i],&num2[i]);
sum[i]=num1[i]+num2[i];
diff[i]=num1[i]-num2[i];
}
memset(dp,-,sizeof(dp)); dp[][fix]=;
for(int i=;i<=m;i++){
for(int k=;k<=*fix;k++){
if(dp[i-][k]<){continue;}
for(int j=;j<=n;j++){
if(dp[i][k+diff[j]]<dp[i-][k]+sum[j]&&check(i-,k,j)){
dp[i][k+diff[j]]=dp[i-][k]+sum[j];
path[i][k+diff[j]]=j;
}
}
}
}
int minn=,maxx=;
for(int i=;i<=fix;i++){
// cout<<dp[m][fix-i];
if(dp[m][fix-i]>=||dp[m][fix+i]>=){
minn=i;break;
}
}
int rec=;
if(dp[m][fix-minn]>dp[m][fix+minn]){
rec=fix-minn;
maxx=dp[m][fix-minn];
}
else{
rec=fix+minn;
maxx=dp[m][fix+minn];
}
int ans1,ans2;
ans1=ans2=;
while(m){
ans.push_back(path[m][rec]);
ans1+=num1[path[m][rec]];
ans2+=num2[path[m][rec]];
rec-=diff[path[m][rec]];
m--;
}
cases++;
printf("Jury #%d\n",cases);
printf("Best jury has value %d for prosecution and value %d for defence:\n",ans1,ans2);
sort(ans.begin(),ans.end());
int sz=ans.size();
for(int i=;i<sz;i++){
printf(" %d",ans[i]);
}
printf("\n\n");
} return ;
}
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