HDU-1003 Max Sum(动态规划，最长字段和问题)
HDU-1003 Max Sum(动态规划，最长字段和问题)
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 193355 Accepted Submission(s): 45045
Problem Description Given a sequence a[1],a[2],a[3]……a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input 2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
Sample Output Case 1: 14 1 4
Case 2: 7 1 6
这是线性动态规划比较简单的最长子段和的问题，状态转移方程 if(dp[i-1]>=0) dp[i]=dp[i-1]+a[i]; else { dp[i]=a[i]; } 这道题目可以用数组，也可以用滚动数组的效果，节省空间、
用一维数组
#include <iostream>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <stdlib.h>
using namespace std;
int n;
int a[100005];
int dp[100005];
int start;
int _end;
int main()
{
int t;
scanf("%d",&t);
for(int cas=1;cas<=t;cas++)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
memset(dp,0,sizeof(dp));
_end=1;
dp[1]=a[1];
for(int i=2;i<=n;i++)
{
if(dp[i-1]>=0)
dp[i]=dp[i-1]+a[i];
else
{
dp[i]=a[i];
}
}
int max=dp[1];
for(int i=2;i<=n;i++)
{
if(max<dp[i])
{
max=dp[i];
_end=i;
}
}
int t1=0;
start=_end;
for(int i=_end;i>0;i--)
{
t1=t1+a[i];
if(t1==max)
start=i;
}
cout<<"Case "<<cas<<":"<<endl<<max<<" "<<start<<" "<<_end<<endl;
if(cas!=t)
printf("\n");
}
return 0;
}滚动数组
#include <iostream>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <stdlib.h>
using namespace std;
int n;
int a;
int sum;
int _begin;
int _end;
int main()
{
int t;
scanf("%d",&t);
int k=0;
while(t--)
{
int max;
int x=1;
scanf("%d%d",&n,&a);
sum=a;
max=a;
_begin=_end=1;
for(int i=2;i<=n;i++)
{
scanf("%d",&a);
if(sum>=0)
{
sum+=a;
}
else
{
sum=a;
x=i;
}
if(max<sum)
{
max=sum;
_begin=x;
_end=i;
}
}
cout<<"Case "<<++k<<":"<<endl<<max<<" "<<_begin<<" "<<_end<<endl;
if(t)
cout<<endl;
}
return 0;
}