Hdu1003 maximum continuous subcolumns and (Dynamic Planning)★★★☆☆)

Solution:

The dynamic programming method is used to solve the maximum subcolumn sum. Corresponding to input data a [I], there will be data DP [I]. Each element DP [I] in the array DP represents the maximum continuous subcolumn and ending with a [I. Traverse the DP array to find the child column and maximum value.

If (DP [I-1]> = 0 ){

DP [I] = DP [I-1] + A [I];

}

Else {

DP [I] = A [I]

}

 

#include <iostream>#include <stdio.h>#include <stdlib.h>using namespace std;int r[100001];int main(){    //freopen("in.txt","r",stdin);    int cs;    r[0] = -1000000;    scanf("%d",&cs);    int n;    for(int t=1;t<=cs;t++){        scanf("%d",&n);        int s=0,e=0,m=-1000000;        int a,b,tmp;        for(int i=1;i<=n;i++){            scanf("%d",&r[i]);            if(r[i-1]>=0){                r[i]+=r[i-1];                b=i;                tmp = r[i];            }            else{                a = b = i;                tmp = r[i];            }            if(tmp>m){                m = tmp;                s = a;                e = b;            }        }        if(t!=1){            printf("\n");        }        printf("Case %d:\n%d %d %d\n",t,m,s,e);    }    //system("pause");    return 0;}