The Blacklist

Problem Statement :

```A new gangster is trying to take control of the city. He makes a list of his N adversaries (e.g. gangster 1, gangster 2, ... gangster N-1, gangster N) and plans to get rid of them.

K mercenaries are willing to do the job. The gangster can use any number of these mercenaries. But he has to honor one condition set by them: they have to be assigned in such a way that they eliminate a consecutive group of gangsters in the list, e.g. gangster i, gangster i+1, ..., gangster j-1, gangster j, where the following is true: 1 <= i <= j <= N.

While our new gangster wants to kill all of them, he also wants to pay the least amount of money. All mercenaries charge a different amount to kill different people. So he asks you to help him minimize his expenses.

Input Format

The first line contains two space-separated integers, N and K. Then K lines follow, each containing N integers as follows:
The jth number on the ith line is the amount charged by the ith mercenary for killing the jth gangster on the list.

Constraints
1 <= N <= 20
1 <= K <= 10
0 <= amount charged <= 10000

Output Format

Just one line, the minimum cost for killing the N gangsters on the list.```

Solution :

```                            ```Solution in C :

In C++ :

#include <cstdio>
#include <cmath>
#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <map>
#include <cassert>
#include <string>
#include <cstring>

using namespace std;

#define rep(i,a,b) for(int i = a; i < b; i++)
#define S(x) scanf("%d",&x)
#define P(x) printf("%d\n",x)

typedef long long int LL;
const int INF = 100000000;
int C[10][20];
int memo[1<<10][11][21];
int n,k;

int solve(int mask, int last, int idx) {
// printf("%d %d %d\n",mask,last,idx);
if(last == -1) {
int res = INF;
rep(i,0,k) res = min(res, C[i][idx] + solve(mask|(1<<i), i, idx+1));

return res;
}
if(idx == n) return 0;

int &res = memo[mask][last][idx];

res = C[last][idx] + min(res, solve(mask, last, idx+1));

res = min(res, C[i][idx] + solve(mask|(1<<i) , i, idx+1) );
}

return res;
}

int main() {
scanf("%d%d",&n,&k);

rep(i,0,1<<k) rep(j,0,k) rep(l,0,n) memo[i][j][l] = INF;

rep(i,0,k) rep(j,0,n) S(C[i][j]);

printf("%d\n",solve(0, -1, 0));

return 0;

}

In Java :

import java.util.Scanner;

public class Solution {

public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int N = input.nextInt();
int K = input.nextInt();
int[][] price = new int[K][N];
for (int k=0; k<K; k++) {
for (int n=0; n<N; n++) {
price[k][n] = input.nextInt();
}
}
int limit = 1<<K;
int[][] dp = new int[N+1][limit+1];
for (int n=0; n<=N; n++) {
for (int x=0; x<=limit; x++) {
dp[n][x] = 1000000;
}
}
dp[0][0] = 0;
for (int n=1; n<=N; n++) {
for (int k=0; k<K; k++) {
int mask = 1 << k;
for (int from=1; from<=n; from++) {
for (int x=0; x<limit; x++) {
if ((x&mask) == 0) {
int newValue = dp[from-1][x];
for (int i=from; i<=n; i++) {
newValue += price[k][i-1];
}
if (dp[n][newMask] > newValue) {
}
}
}
}
}
}
int min = Integer.MAX_VALUE;
for (int value : dp[N]) {
min = Math.min(min, value);
}
System.out.println(min);
}

}

In C :

#include<stdio.h>
#include<limits.h>
#define MIN(x,y) x<y?x:y
int res,n,k,a[15][25]={0};
void find(int curr,int d[],int i,int sumsofar)
{

//printf("here %d %d %d %d\n",curr,i,sumsofar,res);
if(sumsofar>=res)
return;
if(curr>n)
{
res=sumsofar;
return;
}
int p[25],x=0;

//find(curr+1,d,i,sumsofar+a[i][curr]);
int j,l;
for(j=1;j<=k;j++)
if(d[j]==0)
p[x++]=j;
for(l=curr;l<=n;l++)
{
for(j=0;j<x;j++)
if(sumsofar+a[p[j]][l]<sumsofar+a[i][l] && sumsofar+a[p[j]][l]<res)
{
// printf("here also %d %d\n",p[j],i);
d[p[j]]=1;
find(l+1,d,p[j],sumsofar+a[p[j]][l]);
d[p[j]]=0;
}
sumsofar+=a[i][l];
if(sumsofar>=res)
return;
}
res=sumsofar;
}
int main()
{
int i,j;
scanf("%d%d",&n,&k);
for(i=1;i<=k;i++)
for(j=1;j<=n;j++)
scanf("%d",&a[i][j]);
/*for(i=1;i<=k;i++)
{for(j=1;j<=n;j++)
{
b[i][j]=b[i][j-1]+a[i][j];
//if(i!=1)
for(j1=1;j1<=i;j1++)
b[i][j]=MIN(b[i][j],(b[j1][j-1]+a[i][j]));
//  printf("%d ",b[i][j]);
}
// printf("\n");
}*/
//int min=INT_MAX;
/*for(i=1;i<=k;i++)
if(b[i][n]<min)
min=b[i][n];*/
res=INT_MAX;
int d[25]={0};
for(i=1;i<=k;i++)
{
d[i]=1;
find(1,d,i,0);
d[i]=0;
}
printf("%d\n",res);
return 0;
}

In Python3 :

G, M = [int(s) for s in input().split()]

costs = list()
cache = dict()

for i in range(M):
costs.append([int(s) for s in input().split()])

def kill(gn, imercs):
key = str(gn)+str(imercs)
if key in cache:
return cache[key]

mercs = imercs[:]
if gn == 0:
return 0

temp = list()
for m in mercs:
tempm = mercs[:]
tempm.remove(m)
if len(tempm) == 0:
temp.append(sum(costs[m][:gn]))
else:
for c in range(gn - 1, -1, -1):
temp.append(sum(costs[m][c:gn]) + kill(c, tempm))
cache[key] = min(temp)
return min(temp)

print(str(kill(G, [i for i in range(M)])))```
```

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