1.poj 1160
Post Office
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 16455 | Accepted: 8916 |
Description
There is a straight highway with villages alongside the highway. The highway is represented as an integer axis, and the position of each village is identified with a single integer coordinate. There are no two villages in the same position. The distance between two positions is the absolute value of the difference of their integer coordinates.
Post offices will be built in some, but not necessarily all of the villages. A village and the post office in it have the same position. For building the post offices, their positions should be chosen so that the total sum of all distances between each village and its nearest post office is minimum.
You are to write a program which, given the positions of the villages and the number of post offices, computes the least possible sum of all distances between each village and its nearest post office.
Input
Your program is to read from standard input. The first line contains two integers: the first is the number of villages V, 1 <= V <= 300, and the second is the number of post offices P, 1 <= P <= 30, P <= V. The second line contains V integers in increasing order. These V integers are the positions of the villages. For each position X it holds that 1 <= X <= 10000.
Output
The first line contains one integer S, which is the sum of all distances between each village and its nearest post office.
Sample Input
10 5 1 2 3 6 7 9 11 22 44 50
Sample Output
9 dp【i】[j]=dp[k][j-1]+sum[k+1][i];注意sum数组可以进行优化
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
#define INF 0x7fffffff
int v,p,dp[310][31],a[310],sum[310][310];
int main()
{
while(scanf("%d%d",&v,&p)!=EOF)
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=v;i++)
scanf("%d",&a[i]);
for(int i=1;i<=v;i++)
{
for(int j=i+1;j<=v;j++)
{
sum[i][j]=sum[i][j-1]+a[j]-a[(i+j)/2];
}
}
for(int i=1;i<=v;i++)
{
dp[i][1]=sum[1][i];
}
for(int j=2;j<=p;j++)
{
for(int i=j+1;i<=v;i++)
{
dp[i][j]=INF;
for(int k=j;k<=i;k++)
dp[i][j]=min(dp[i][j],dp[k][j-1]+sum[k+1][i]);
}
}
printf("%d\n",dp[v][p]);
}
return 0;
}
2.poj 1050
To the Max
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 42088 | Accepted: 22375 |
Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:
9 2
-4 1
-1 8
and has a sum of 15.
Input
The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].
Output
Output the sum of the maximal sub-rectangle.
Sample Input
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
Sample Output
15 相当于最大连续子序列和而已，水。。。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
using namespace std;
int a[101][101],n,cal[101][101],sum,maxx;
int main()
{
while(scanf("%d",&n)!=EOF)
{
sum=0,maxx=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
scanf("%d",&a[i][j]);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
cal[i][j]=cal[i][j-1]+a[i][j];
}
}
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
sum=0;
for(int k=1;k<=n;k++)
{
sum+=cal[k][j]-cal[k][i-1];
if(sum>maxx)
maxx=sum;
else if(sum<0)
sum=0;
}
}
}
printf("%d\n",maxx);
}
return 0;
}