Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 27582 Accepted Submission(s): 9617
Problem Description
Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given a consecutive number sequence S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (1 ≤ i ≤ j ≤ n). Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x is not allowed). But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S 1, S 2, S 3 ... S n. Process to the end of file.
Output
Output the maximal summation described above in one line.
Sample Input
1 3 1 2 3
2 6 -1 4 -2 3 -2 3
Sample Output
6
8
Hint
Huge input, scanf and dynamic programming is recommended. 题意：
求最大m子段和。
代码：
//最大m子段和递推公式:dp[i][j]=max(dp[i][j-1]+num[j],dp[i-1][t]+num[j]) (i<=t<=j-1)//优化：因为每次都要求一个最大的dp[i-1][t](i<=t<=j-1),可以每次求出来dp[i][j]时保存//这个值，到下一次时直接用就行了，这样每次就只用到了dp[i][j]和dp[i][j-1],可以省去一维。#include#include #include using namespace std;const int maxn=1000006;const int inf=0x7fffffff;int dp[maxn],fdp[maxn],num[maxn];int main(){ int m,n,maxnum; while(scanf("%d%d",&m,&n)==2){ for(int i=1;i<=n;i++) scanf("%d",&num[i]); memset(dp,0,sizeof(dp)); memset(fdp,0,sizeof(fdp)); for(int i=1;i<=m;i++){ maxnum=-inf; for(int j=i;j<=n;j++){ dp[j]=max(dp[j-1]+num[j],fdp[j-1]+num[j]); fdp[j-1]=maxnum; maxnum=max(maxnum,dp[j]); } } printf("%d\n",maxnum); } return 0;}