HDU-1024-Max Sum Plus Plus
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题目网址:Max Sum Plus Plus
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)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 S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (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(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (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(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx 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(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
Process to the end of file.
Process to the end of file.
Output
Output the maximal summation described above in one line.
Sample Input
1 3 1 2 32 6 -1 4 -2 3 -2 3
Sample Output
68HintHuge input, scanf and dynamic programming is recommended.
Author
JGShining(极光炫影)
Recommend
思路:dp[i][j]表示前j项含有i个子段的最大和,那么dp[i][j]可以由哪些状态的到呢,首先有dp[i][j-1]+num[j],j-1个元素分为i段的最大值+num[j],这种情况num[j]属于第i个子段也就是原来的第i个子段加了一个num[j]元素,然后是dp[i-1][j-1]+num[j],j-1个元素分为i-1段的最大值+num[j],这种情况num[j]自己成为新的一段即第i段。所以就得到了动态转移方程
动态转移方程:dp[i][j]=max(dp[i][j-1]+num[j],dp(i-1,t)+num[j]),其中i-1<=t<=j-1
由于题目数据量太大(m*n)我们开不了这么大的数组,所以我们得对空间进压缩,可以发现我们只用到两个时候的dp数组即dp[i][j]和dp[i-1][t],i-1<=t<=j-1而i-1之前的用过了都用不到了,所以我们可以开两个一维数组dp表示分为i段的状态,Max[j-1]表示i-1到j-1元素分为i-1段的状态的最大值。
然后还有一个很巧妙的地方就是Max数组的更新,当j元素添加到了第i段,等到推j元素分成i+1段时,Max数组也要跟着i更新。
卡了好长时间的一道题,一直想不明白,后来才明白,其中那个Max用的很巧妙,大家理解理解。
#include<bits/stdc++.h>using namespace std; const int maxn=1000005;const int INF=1<<29;int n,m,num[maxn],dp[maxn],pre[maxn],Max;int main() { while(~scanf("%d%d",&m,&n)) { for(int i=1;i<=n;i++) { scanf("%d",&num[i]); } memset(dp,0,sizeof(dp));memset(pre,0,sizeof(pre)); for(int i=1;i<=m;i++) { Max=-INF; for(int j=i;j<=n;j++) { dp[j]=max(dp[j-1]+num[j],pre[j-1]+num[j]); pre[j-1]=Max; Max=max(dp[j],Max);}} printf("%d\n",Max); } return 0; }
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