hdu1024 Max Sum Plus Plus dp
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Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24379 Accepted Submission(s): 8377
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.
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状态dp[i][j]
有前j个数,组成i组的和的最大值。
决策: 第j个数,是在第包含在第i组里面,还是自己独立成组。
方程 dp[i][j]=Max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j
空间复杂度,m未知,n<=1000000, 继续滚动数组。
时间复杂度 n^3. n<=1000000. 显然会超时,继续优化。
max( dp[i-1][k] ) 就是上一组 0....j-1 的最大值。我们可以在每次计算dp[i][j]的时候记录下前j个
的最大值 用数组保存下来 下次计算的时候可以用,这样时间复杂度为 n^2.
/*状态dp[i][j]有前j个数,组成i组的和的最大值。决策: 第j个数,是在第包含在第i组里面,还是自己独立成组。方程 dp[i][j]=Max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j空间复杂度,m未知,n<=1000000, 继续滚动数组。 时间复杂度 n^3. n<=1000000. 显然会超时,继续优化。max( dp[i-1][k] ) 就是上一组 0....j-1 的最大值。我们可以在每次计算dp[i][j]的时候记录下前j个的最大值 用数组保存下来 下次计算的时候可以用,这样时间复杂度为 n^2.*/#include<stdio.h>#include<algorithm>#include<iostream>using namespace std;#define MAXN 1000000#define INF 0x7fffffffint dp[MAXN+10];int mmax[MAXN+10];int a[MAXN+10];int main(){ int n,m; int i,j,mmmax; while(scanf("%d%d",&m,&n)!=EOF) { for(i=1;i<=n;i++) { scanf("%d",&a[i]); mmax[i]=0; dp[i]=0; } dp[0]=0; mmax[0]=0; for(i=1;i<=m;i++) { mmmax=-INF; for(j=i;j<=n;j++) { dp[j]=max(dp[j-1]+a[j],mmax[j-1]+a[j]); mmax[j-1]=mmmax; mmmax=max(mmmax,dp[j]); } } printf("%d\n",mmmax); } return 0; }
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