HDU 1024 最大m个子段和滚动数组
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 22175 Accepted Submission(s): 7446
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. ^_^
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.
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 8HintHuge input, scanf and dynamic programming is recommended. 题意:让求从n个数中找出m个子段 的最大和，（每个子段自己是连续的,每个子段之间没有重复元素）设输入的数组为a[1...n]，从中找出m个段，使者几个段的和为最大
dp[i][j]表示前j个数中取i个段的和的最大值，其中最后一个段包含a[j]。
则状态转移方程为：
dp[i][j]=max{dp[i][j-1]+a[j],max{dp[i-1][t]}+a[j]} i-1=<t<j-1
因为dp[i][j]中a[j]可能就自身一个数组成最后一段，或者a[j]与a[j-1]等前面的数组成最后一段。
此题n数据太大，二维数组开不下，而且三重循环，想到状态转移方程后还是困难重重。
想想，二维数组不行的话，肯定要压缩成一维数组：
因为dp[i-1][t]的值只在计算dp[i][j]的时候用到，那么没有必要保存所有的dp[i][j] for i=1 to m,这样我们可以用一维数组存储。
用pre[j]表示j之前一个状态dp[i-1][]中1-j之间，不一定包含a[j]的最大字段和，然后推dp[i][j]状态时，dp[i][j]=max{pre[j-1]，dp[j-1]}+a[j];
褐色的为了方便理解，其实不存在。
4 6
2 -4 5 6 -8 10
dp[1]=2 pre[0]=-100000000
dp[2]=-2 pre[1]=2
dp[3]=5 pre[2]=2
dp[4]=11 pre[3]=5
dp[5]=3 pre[4]=11
dp[6]=13 pre[5]=11
dp[2]=-2 pre[1]=-100000000
dp[3]=7 pre[2]=-2
dp[4]=13 pre[3]=7
dp[5]=5 pre[4]=13
dp[6]=21 pre[5]=13
dp[3]=3 pre[2]=-100000000
dp[4]=13 pre[3]=3
dp[5]=5 pre[4]=13
dp[6]=23 pre[5]=13
dp[4]=9 pre[3]=-100000000
dp[5]=5 pre[4]=9
dp[6]=23 pre[5]=9
23
dp[j]不一定是该行最大的
dp[j]表示将a[j]加上的最大值
#include<bits/stdc++.h> using namespace std; int dp[1000010],a[1000010]; int M[1000010]; //全部存的是 i-1段的最大值 int main() { int n,m; while(~scanf("%d%d",&m,&n)) { for(int i=1;i<=n;i++) scanf("%d",&a[i]); memset(dp,0,sizeof(dp)); memset(M,0,sizeof(M)); int Max=-100000000; for(int i=1;i<=m;i++) { Max=-100000000; for(int j=i;j<=n;j++) //必须从i开始，否则上一轮的最大值会被覆盖掉 { //dp[j]的取值是从 1在第i个子段后面加上a[j] 2是让a[j]成为第i段的第一个元素(独立成为yi段) dp[j]=max(dp[j-1]+a[j],M[j-1]+a[j]); M[j-1]=Max; //用于存放在 j-1个数字中 i-1个子段和的最大值 //printf("dp[%d]=%d pre[%d]=%d\n",j,dp[j],j-1,pre[j-1]); Max=max(Max,dp[j]); } } printf("%d\n",Max); } return 0; }
二维思路
while(~scanf("%d%d",&m,&n)) { for(int i=1;i<=n;i++) scanf("%d",&a[i]); memset(dp,0,sizeof(dp)); memset(M,0,sizeof(M)); int Max=-100000000; for(int i=1;i<=m;i++) { // Max=-100000000; for(int j=i;j<=n;j++) { //dp[j]的取值是从 1在第i个子段后面加上a[j] 2是让a[j]成为第i段的第一个元素(独立成为yi段) if(j>i) { dp[i][j]=dp[i][j-1]+a[j]; for(int k=i-1;k<=j-1;k++) dp[i][j]=max(dp[i][j],dp[i-1][k]+a[j]); } else dp[i][j]=dp[i-1][j-1]+a[j]; } } for(int i=m;i<=n;i++) printf("%d\n",dp[m][i]); }