hdu1024Max Sum Plus Plus【状态dp 滚动数组】
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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
68
1003的升级版==就像之前看1505city game 和 1506Largest Rectangle in a Histogram 没看出来二者有什么关系一样 ==
这次做完了1003maxsum 依旧没做出来1024更何况1003的写法就不对→_→
状态转移方程 dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k])+a[j]) 前者是与之前的成为一组 后者是独立成组
dp[i][j]表示前j个数组成i组 而dp[i-1][k]是在j=j-1时的mmmax (怪不得他的序号就是j-1==)即mmax[j-1]
而mmax[]只有在下一次i=i+1 j=j 这个对应位置会被用到 恰恰说明了dp[i-1][k] 的本质是为了取上一次循环中的最大值
#include <iostream>#include<cstdio>#include<cstring>using namespace std;int a[1000005],dp[1000005],mmax[1000005],n,m,mmmax;int main(){ while(~scanf("%d%d",&m,&n)) { for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } memset(dp,0,sizeof(dp)); memset(mmax,0,sizeof(mmax)); for(int i=1;i<=m;i++) { mmmax=-0x3f3f3f3f; for(int j=i;j<=n;j++) { dp[j]=max(dp[j-1]+a[j],mmax[j-1]+a[j]); mmax[j-1]=mmmax; mmmax=max(dp[j],mmmax); } } printf("%d\n",mmmax); } return 0;}
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