hdu4976 A simple greedy problem.

Posted 2020-09-11 神犇(shenben)

A simple greedy problem.

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 476    Accepted Submission(s): 193

Problem Description

Victor and Dragon are playing DotA. Bored of normal games, Victor challenged Dragon with a competition of creep score (CS). In this competition, there are N enemy creeps for them. They hit the enemy one after another and Dragon takes his turn first. Victor uses a strong melee character so that in his turn, he will deal 1 damage to all creeps. Dragon uses a flexible ranged character and in his turn, he can choose at most one creep and deal 1 damage. If a creep take 1 damage, its health will reduce by 1. If a creep’s current health hits zero, it dies immediately and the one dealt that damage will get one score. Given the current health of each creep, Dragon wants to know the maximum CS he can get. Could you help him?

 

 

Input

The first line of input contains only one integer T(<=70), the number of test cases. 

For each case, the first line contains 1 integer, N(<=1000), indicating the number of creeps. The next line contain N integers, representing the current health of each creep(<=1000).

 

 

Output

Each output should occupy one line. Each line should start with "Case #i: ", with i implying the case number. For each case, just output the maximum CS Dragon can get.

 

 

Sample Input

2 5 1 2 3 4 5 5 5 5 5 5 5

 

 

Sample Output

Case #1: 5 Case #2: 2

 

 

Author

BJTU

 

 

Source

2014 Multi-University Training Contest 10

 

 

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1005;
int T,n,cas,a[N],b[N];
int f[N][N];
int main(){
    for(scanf("%d",&T);T--;){
        scanf("%d",&n);memset(b,-1,sizeof b);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        sort(a+1,a+n+1);
        for(int i=1;i<=n;i++){
            for(int j=a[i];j;j--){
                if(b[j]==-1){
                    b[j]=a[i];break;
                }
            }
        }
        memset(f,-1,sizeof f);
        f[0][0]=0;
        for(int i=0,t;i<a[n];i++){
            for(int j=0;j<=i+1;j++){
                f[i+1][j+1]=f[i][j];
                if((~b[i+1])&&(t=j-b[i+1]+i+1)>=0)f[i+1][t]=max(f[i+1][t],f[i][j]+1);
            }
        }
        int ans=0;
        for(int i=0;i<=a[n];i++){
            ans=max(ans,f[a[n]][i]);
        }
        printf("Case #%d: %d\\n",++cas,ans);
    }
    return 0;
}