Codeforces 467C George and Job(DP)
2016-08-14 22:19
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题目
Source
http://codeforces.com/contest/467/problem/CDescription
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.
Input
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).Output
Print an integer in a single line — the maximum possible value of sum.Sample Input
5 2 11 2 3 4 5
7 1 3
2 10 7 18 5 33 0
Sample Output
961
分析
题目大概说给一个序列,求k个不重叠长m的连续子序列的最大和。DP搞了。
dp[i][j]表示前i个数中构成j个子序列的最大和
转移就是dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+sum[i]-sum[i-m]),sum是前缀和
代码
#include<cstdio> #include<algorithm> using namespace std; long long d[5555][5555],sum[5555]; int a[5555]; int main(){ int n,m,k; scanf("%d%d%d",&n,&m,&k); for(int i=1; i<=n; ++i) scanf("%d",a+i),sum[i]=sum[i-1]+a[i]; for(int i=1; i<=n; ++i){ for(int j=1; j<=k; ++j){ if(j*m>i) break; d[i][j]=max(d[i-1][j],d[i-m][j-1]+sum[i]-sum[i-m]); } } printf("%lld",d [k]); return 0; }
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