Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14654 Accepted Submission(s): 4778
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 ≤ jxor 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 ≤ jxor 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 3 2 6 -1 4 -2 3 -2 3
Sample Output
6 8HintHuge input, scanf and dynamic programming is recommended.
题目大意：设输入的数组为a[1...n]，从中找出m个连续的段，使得这几个段的和为最大。
解题思路：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]。当然我们需要开的数组可能会爆内存，dp[10^6][m]，m给多大我们是不知道的。如果是10的话根本开不了。我们先找到状态转移方程：
dp[i][j]=max{dp[i][j-1],max{dp[i-1][t]}} +a[j] 1=<t<=j-1
我们计算dp[i][j]的时候只会用到dp[i-1][t]中的最大值，因此我们可以用一个last数组将他们保存。
于是状态转移方程转化为：
dp[j]=max(dp[j-1],last[j-1])+a[j].
当然last需要随时更新。
但是分析数据
应该是要用long long的？最大值是3*10^10，但是会超时，纠结了一个小时
求路过的大神帮忙看看。。
题目地址：Max Sum Plus Plus
AC代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1000005;
//dp[i][j]=max(dp[i][j-1],dp[i-1][t])+a[j] (1<=t<=j-1)
int a[maxn];
int dp[maxn];
int last[maxn];
int main()
{
int n,m;
int i,j;
while(~scanf("%d%d",&m,&n))
{
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(i=0;i<=1e6;i++)
dp[i]=last[i]=0;
int ma;
for(i=1;i<=m;i++)
{
ma=-1e9; //更新dp[i-1][t](1<=t<=j-1)即last的最大值
for(j=i;j<=n;j++)
{
if(dp[j-1]>last[j-1]) dp[j]=dp[j-1]+a[j];
else dp[j]=last[j-1]+a[j];
last[j-1]=ma;
if(dp[j]>ma) ma=dp[j];
}
}
printf("%d\n",ma);
}
return 0;
}
/*
1 3 1 2 3
2 6 -1 4 -2 3 -2 3
*/