Codeforces Round #267 (Div. 2) C. George and Job (dp)
wa哭了,,t哭了,,还是看了题解。。。
8170436 | 2014-10-11 06:41:51 | njczy2010 | C - George and Job | GNU C++ | Accepted | 109 ms | 196172 KB |
8170430 | 2014-10-11 06:39:47 | njczy2010 | C - George and Job | GNU C++ | Wrong answer on test 1 | 61 ms | 196200 KB |
8170420 | 2014-10-11 06:37:14 | njczy2010 | C - George and Job | GNU C++ | Runtime error on test 23 | 140 ms | 196200 KB |
8170348 | 2014-10-11 06:16:59 | njczy2010 | C - George and Job | GNU C++ | Time limit exceeded on test 11 | 1000 ms | 196200 KB |
8170304 | 2014-10-11 06:05:15 | njczy2010 | C - George and Job | GNU C++ | Time limit exceeded on test 12 | 1000 ms | 196200 KB |
8170271 | 2014-10-11 05:53:29 | njczy2010 | C - George and Job | GNU C++ | Wrong answer on test 3 | 77 ms | 196200 KB |
8170266 | 2014-10-11 05:52:37 | njczy2010 | C - George and Job | GNU C++ | Wrong answer on test 3 | 62 ms | 196200 KB |
8170223 | 2014-10-11 05:39:00 | njczy2010 | C - George and Job | GNU C++ | Time limit exceeded on test 12 | 1000 ms | 196100 KB |
8170135 | 2014-10-11 05:07:06 | njczy2010 | C - George and Job | GNU C++ | Wrong answer on test 9 | 93 ms | 196100 KB |
8170120 | 2014-10-11 05:01:28 | njczy2010 | C - George and Job | GNU C++ | Wrong answer on test 5 | 78 ms | 196100 KB |
1 second
256 megabytes
standard input
standard output
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.
Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers:
[l1, r1], [l2, r2], ..., [lk, rk] (1 ≤ l1 ≤ r1 < l2 ≤ r2 < ... < lk ≤ rk ≤ n; ri - li + 1 = m),
in such a way that the value of sum is maximal possible. Help George to cope with the task.
The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109).
Print an integer in a single line — the maximum possible value of sum.
5 2 1 1 2 3 4 5
9
7 1 3 2 10 7 18 5 33 0
61
转移方程为
dp[i][j]=max(dp[i+1][j],dp[i+m][j-1]+sum[i]); (转自:http://www.tuicool.com/articles/m6v6z23) C:显然是dp,设f[i][j]表示第i段的结尾为j时的最优值,显然f[i][j]=max{f[i-1][k]+sum[j]-sum[j-m]}(0<=k<=j-m) (转自:http://www.bkjia.com/ASPjc/881176.html)
不过这样是O(k*n^2),可能超时。
我们发现每一阶段的转移能用到的最优状态都是上一阶段的前缀最优值,于是dp时直接记录下来用来下面的转移,这样就不用枚举了,变为O(k*n)水过。
貌似卡内存,用了滚动数组。
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<string>
//#include<pair> #define N 5005
#define M 1000005
#define mod 1000000007
//#define p 10000007
#define mod2 100000000
#define ll long long
#define LL long long
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b) using namespace std; int n,m,k;
ll dp[N][N];
ll p[N];
ll ans;
ll sum[N]; void ini()
{
memset(dp,,sizeof(dp));
memset(sum,,sizeof(sum));
int i;
ll te=;
for(i=;i<=n;i++){
scanf("%I64d",&p[i]);
} for(i=n-m+;i<=n;i++){
te+=p[i];
}
sum[n-m+]=te;
dp[n-m+][]=te;
// ma=te;
for(i=n-m;i>=;i--){
sum[i]=sum[i+]+p[i]-p[i+m];
}
//for(i=1;i<=n;i++) printf(" i=%d sum=%I64d\n",i,sum[i]);
} void solve()
{
int i,j;
dp[n][]=sum[n];
for(i=n-;i>=;i--){
for(j=k;j>=;j--){
if(i+m<=n+)
dp[i][j]=max(dp[i+][j],dp[i+m][j-]+sum[i]);
else
dp[i][j]=dp[i+][j];
}
} } void out()
{
//for(int i=1;i<=n;i++){
// for(int j=0;j<=k;j++){
// printf(" i=%d j=%d dp=%I64d\n",i,j,DP(i,j));
// }
// }
printf("%I64d\n",dp[][k]);
} int main()
{
// freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
// scanf("%d",&T);
// for(int ccnt=1;ccnt<=T;ccnt++)
// while(T--)
while(scanf("%d%d%d",&n,&m,&k)!=EOF)
{
//if(n==0 && k==0 ) break;
//printf("Case %d: ",ccnt);
ini();
solve();
out();
} return ;
}
Codeforces Round #267 (Div. 2) C. George and Job (dp)的更多相关文章
- Codeforces Round #267 (Div. 2) C. George and Job(DP)补题
Codeforces Round #267 (Div. 2) C. George and Job题目链接请点击~ The new ITone 6 has been released recently ...
- Codeforces Round #390 (Div. 2) C. Vladik and chat(dp)
http://codeforces.com/contest/754/problem/C C. Vladik and chat time limit per test 2 seconds memory ...
- Codeforces Round #605 (Div. 3) D. Remove One Element(DP)
链接: https://codeforces.com/contest/1272/problem/D 题意: You are given an array a consisting of n integ ...
- Codeforces Round #651 (Div. 2) E. Binary Subsequence Rotation(dp)
题目链接:https://codeforces.com/contest/1370/problem/E 题意 给出两个长为 $n$ 的 $01$ 串 $s$ 和 $t$,每次可以选择 $s$ 的一些下标 ...
- Codeforces Round #321 (Div. 2) D Kefa and Dishes(dp)
用spfa,和dp是一样的.转移只和最后一个吃的dish和吃了哪些有关. 把松弛改成变长.因为是DAG,所以一定没环.操作最多有84934656,514ms跑过,实际远远没这么多. 脑补过一下费用流, ...
- 01背包 Codeforces Round #267 (Div. 2) C. George and Job
题目传送门 /* 题意:选择k个m长的区间,使得总和最大 01背包:dp[i][j] 表示在i的位置选或不选[i-m+1, i]这个区间,当它是第j个区间. 01背包思想,状态转移方程:dp[i][j ...
- Codeforces Round #184 (Div. 2) E. Playing with String(博弈)
题目大意 两个人轮流在一个字符串上删掉一个字符,没有字符可删的人输掉游戏 删字符的规则如下: 1. 每次从一个字符串中选取一个字符,它是一个长度至少为 3 的奇回文串的中心 2. 删掉该字符,同时,他 ...
- Codeforces Round #556 (Div. 2) - C. Prefix Sum Primes(思维)
Problem Codeforces Round #556 (Div. 2) - D. Three Religions Time Limit: 1000 mSec Problem Descripti ...
- Codeforces Round #394 (Div. 2) E. Dasha and Puzzle(分形)
E. Dasha and Puzzle time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
随机推荐
- access处理重复创建表的方法。
第一种,使用MSysObjects表查找表名为当前创建表的名字的内容,相当于普通查询,但是access数据库有一个安全问题,就是有时候一开始是没有权限去调这些系统表的,这时可以再2007的access ...
- idea报错:The server time zone value '�й���ʱ��' is unrecognized or represents more than one time zone. You must configure either the server or JDBC driver (via the serverTimezone configu
java.sql.SQLException: The server time zone value '�й���ʱ��' is unrecognized or represents more tha ...
- 改变console.log的输出样式
console.log允许你通过css来格式化输出,格式如下: console.log(‘%c字符串%c字符串’, 样式1, [样式2]) 其中”%c”为模板字符串 例子: 1 console.log ...
- tcpdump简单使用
1.使用wincap将文件放入系统任意路径, 2.进入系统,赋文件可执行权限, 3.输入命令:./tcpdump -i eth0 -s 0 -w xxx.pcap 4.进行数据交互 5.退出程序运行, ...
- python:加密模块
加密:import hashlib # import md5 #python2 中可以直接引入md5,3中没有 #md5 #md5加密是不可逆的,即不能解密. #只要用MD5加密,结果都是一样的,不区 ...
- Linux内核漏洞利用-环境配置(转)
实验环境: Ubuntu-14.04.1 x86 linux-2.6.32.1 busybox-1.27.2 qemu 0x00 安装qemu sudo apt-get install qemu qe ...
- Php教程
第一部:PHP基础部分(131集,发布完毕) 讲html与PHPt基础,PHP环境搭建,与留言本编写. 下载地址:① HTML视频[2014新版] http://pan.baidu.com/s/1ve ...
- java第十次作业:oop的第6张图片到第11张图片
- xhEditor编辑器上传图片到 OSS
前段时间,公司在项目上用到了xhEditor编辑器来给用户做一个上传图片的功能当时做的时候觉得很有意思,想想 基本的用户图片上传到自己服务器,还有点小占地方: 后来....然后直接上传到阿里云 .接下 ...
- 【css】【动画】【转发】旋转动画
<!DOCTYPE HTML> <html> <head> <meta charset="utf-8"> <s ...