A - Max Sum Plus Plus——POJ

A - Max Sum Plus Plus

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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. ^_^ 

 

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. 

 

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 

Hint

 Huge input, scanf and dynamic programming is recommended.          

 

动规方程：

dp[i][j]=前j个数选择i段的最大值

dp[i][j]=max(dp[i][j-1]  ,  max{dp[i-1][k]}  )+a[j]  (i-1<=k<=j-1)

i独立成段：

其不独立成段，既与j-2成段

#include<cstdio>#include<algorithm>#include<cmath>#include<cstdlib>#include<cstring>#include<vector>using namespace std;const int maxn=1000005;typedef __int64 LL;int a[maxn],dp[maxn];int pre[maxn];//对应递推式的第二项int main()

{    int T;    //freopen("Text//in.txt","r",stdin);    int n,m;    while(~scanf("%d%d",&m,&n))    {        for(int i=1;i<=n;i++){            scanf("%d",a+i);            dp[i]=pre[i]=0;        }        int mx;        dp[0]=pre[0]=0;        for(int i=1;i<=m;i++){            mx=-999999999;            for(int j=i;j<=n;j++)    {                dp[j]=max(dp[j-1],pre[j-1])+a[j];                pre[j-1]=mx;                mx=max(dp[j],mx);            }        }        printf("%d\n",mx);    }    return 0;}

 for(int j=i;j<=n;j++)
{
                dp[j]=max(dp[j-1],pre[j-1])+a[j];
                pre[j-1]=mx;
                mx=max(dp[j],mx);
}

蓝色的表示其最大值

下面我来解释其具体怎么实现：

利用了动态数组，只保留了当前行和上一行，因为动规方程只用的到这两行，

pre【i】为上上一行i之前的最大值

而dp【i】则保留当前行

 for(int j=i;j<=n;j++)
{
                dp[j]=max(dp[j-1],pre[j-1])+a[j];
                pre[j-1]=mx;
                mx=max(dp[j],mx);
}

先利用上一行的j-1数据用于动规方程，

然后把pre[j-1]=mx赋值为当前行的前[j-1]行最小值

这样就保证了,pre[]向下循环

一下代码用于测试;

#include<cstdio>#include<algorithm>#include<cmath>#include<cstdlib>#include<cstring>#include<vector>using namespace std;const int maxn=1000005;typedef __int64 LL;int a[maxn],dp[maxn];int pre[maxn];//对应递推式的第二项int main(){    int T;    //freopen("Text//in.txt","r",stdin);    int n,m;    while(~scanf("%d%d",&m,&n)){        for(int i=1;i<=n;i++){            scanf("%d",a+i);            dp[i]=pre[i]=0;        }        int mx;        dp[0]=pre[0]=0;        for(int i=1;i<=m;i++){            mx=-99999;            for(int j=i;j<=n;j++){                dp[j]=max(dp[j-1],pre[j-1])+a[j];                pre[j-1]=mx;                mx=max(dp[j],mx);                printf("\ndp:\n");                for(int k=1;k<n;k++)                printf("%d ",dp[k]);                printf("\npre\n");                for(int k=1;k<n;k++)                printf("%d ",pre[k]);                printf("\n");            }            printf("\n*********************************\n");        }        printf("%d\n",mx);    }    return 0;}

我把它写简明些：

#include<cstdio>#include<algorithm>#include<cmath>#include<cstdlib>#include<cstring>#include<vector>using namespace std;const int maxn=1000005;typedef __int64 LL;int a[maxn],dp[maxn];int pre[maxn];//对应递推式的第二项int main(){    int T;    //freopen("Text//in.txt","r",stdin);    int n,m;    while(~scanf("%d%d",&m,&n)){        for(int i=1;i<=n;i++){            scanf("%d",a+i);            dp[i]=pre[i]=0;        }        int mx;        dp[0]=pre[0]=0;        for(int i=1;i<=m;i++){mx=-99999;//一定是最小，因为当i==j时，其值一定加上a[j]             for(int j=i;j<=n;j++)//dp[]中储存当前行，和以前行第一个数 {                dp[j]=max(dp[j-1],pre[j-1])+a[j];//先用                 pre[j-1]=mx;//后覆盖                 mx=max(dp[j],mx);//求用于覆盖的最大数                             }        }        printf("%d\n",mx);    }    return 0;}