首页 > 代码库 > hdu4976 A simple greedy problem.
hdu4976 A simple greedy problem.
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).
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
251 2 3 4 555 5 5 5 5
Sample Output
Case #1: 5Case #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;}
hdu4976 A simple greedy problem.
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。