HDU Max Sum Plus Plus DP

Max Sum Plus PlusTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 26559    Accepted Submission(s): 9230Problem DescriptionNow 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. ^_^InputEach test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.Process to the end of file.OutputOutput the maximal summation described above in one line.Sample Input1 3 1 2 32 6 -1 4 -2 3 -2 3Sample Output68HintHuge input, scanf and dynamic programming is recommended.AuthorJGShining（极光炫影）

问题很简单 就是n个数 求最大m段和
很久前就看到过这题 当时因为m没给范围 考虑了下复杂度就没做了
真正动手试试 其实这题数据并不大 O(m*n)也是可以AC的

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d[i][j][0] = 前i个数中 选出j段 而且 第i个数没有被用上d[i][j][1] = 前i个数中 选出j段 而且 第i个数被用上了d[i][j][0] = max(d[i-1][j][0],d[i-1][j][1]);d[i][j][1] = max(d[i-1][j-1][0],d[i-1][j-1][1],d[i-1][j][1]);d[0][0][0]=0;d[0][0][1]=-inf;d[0][j][k]=-inf;   //j>0 k=0,1

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#include<iostream>#include<stdlib.h>#include<stdio.h>#include<string>#include<vector>#include<deque>#include<queue>#include<algorithm>#include<set>#include<map>#include<stack>#include<time.h>#include<math.h>#include<list>#include<cstring>#include<fstream>//#include<memory.h>using namespace std;#define ll long long#define ull unsigned long long#define pii pair<int,int>#define INF 1000000007#define pll pair<ll,ll>#define pid pair<int,double>const int N=1000000+5;int elem[N];const ll inf=1e18;ll d[2][N][2];ll dp(int m,int n){    ll(*d1)[2]=d[0];    ll(*d2)[2]=d[1];    for(int i=0;i<=n;++i){        d1[i][0]=d1[i][1]=-inf;    }    d1[0][0]=0;    for(int i=1;i<=n;++i){        for(int j=1;j<=m;++j){            d2[j][0]=max(d1[j][0],d1[j][1]);            d2[j][1]=max(d1[j-1][0]+elem[i],                    max(d1[j][1]+elem[i],d1[j-1][1]+elem[i]));        }        swap(d1,d2);    }    return max(d1[m][0],d1[m][1]);}int main(){    //freopen("/home/lu/文档/r.txt","r",stdin);    //freopen("/home/lu/文档/w.txt","w",stdout);    int m,n;    while(~scanf("%d%d",&m,&n)){        for(int i=1;i<=n;++i){            scanf("%d",&elem[i]);        }        printf("%lld\n",dp(m,n));    }    return 0;}