今天刷了两道区间dp的进阶题,再次写下题解。
1. HDOJ 2513 Cake slicing
Cake slicing
Time Limit: 2000/1000 MS (Java/Others) Memory Limit:32768/32768 K (Java/Others)
Total Submission(s): 283 Accepted Submission(s): 139
Problem Description
A rectangular cake with a grid of m*n unit squares on its top needs to be slicedinto pieces. Several cherries are scattered on the top of the cake with at mostone cherry on a unit square. The slicing should follow the rules below:
1. each piece is rectangular or square;
2. each cutting edge is straight and along a grid line;
3. each piece has only one cherry on it;
4. each cut must split the cake you currently cut two separate parts
For example, assume that the cake has a grid of 3*4 unit squares on its top,and there are three cherries on the top, as shown in the figure below.
One allowable slicing is as follows.
For this way of slicing , the total length of the cutting edges is 2+4=6.
Another way of slicing is
In this case, the total length of the cutting edges is 3+2=5.
Give the shape of the cake and the scatter of the cherries , you are supposedto find
out the least total length of the cutting edges.
Input
The input file contains multiple test cases. For each test case:
The first line contains three integers , n, m and k (1≤n, m≤20), where n*m isthe size of the unit square with a cherry on it . The two integers showrespectively the row number and the column number of the unit square in thegrid .
All integers in each line should be separated by blanks.
Output
Output an integer indicating the least total length of the cutting edges.
Sample Input
3 4 3
1 2
2 3
3 2
Sample Output
Case 1: 5
分析:
用dp[i][j][k][l]表示以(i,j)为左上角,(k,l)为右下角的矩形切成每份一个樱桃的最小切割长度。然后就利用区间DP的性质,枚举切割点,从小区间转移到大区间。由于这道题不同区间樱桃个数不同,故用递归的写法会方便些。
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
using namespace std;
int dp[25][25][25][25];
bool vis[25][25];
int fun(int a,int b,int c,int d)
{
if(dp[a][b][c][d]!=-1)
return dp[a][b][c][d];
int cnt=0,mi;
for(int i=a;i<=c;i++)
for(int j=b;j<=d;j++)
if(vis[i][j]) cnt++;
if(cnt<=1)
return dp[a][b][c][d]=0;
for(int i=a;i<c;i++)
{
mi=min(mi,fun(a,b,i,d)+fun(i+1,b,c,d)+d-b+1);
}
for(int i=b;i<d;i++)
{
mi=min(mi,fun(a,b,c,i)+fun(a,i+1,c,d)+(c-a+1));
}
return dp[a][b][c][d]=mi;
}
int main()
{
int m,n,k,t=0;
while(~scanf("%d%d%d",&n,&m,&k))
{
t++;
memset(vis,0,sizeof(vis));
memset(dp,-1,sizeof(dp));
for(int i=1;i<=k;i++)
{
int a,b;
scanf("%d%d",&a,&b);
vis[a][b]=1;
}
int ans=fun(1,1,n,m);
printf("Case %d: %d\n",t,ans);
}
return 0;
}
String painter
Time Limit: 5000/2000 MS (Java/Others) Memory Limit:32768/32768 K (Java/Others)
Total Submission(s): 4691 Accepted Submission(s): 2208
Problem Description
There are two strings A and B with equal length. Both strings are made up of lowercase letters. Now you have a powerful string painter. With the help of thepainter, you can change a segment of characters of a string to any othercharacter you want. That is, after using the painter, the segment is made up ofonly one kind of character. Now your task is to change A to B using stringpainter. What’s the minimum number of operations?
Input
Input contains multiple cases. Each case consists of two lines:
The first line contains string A.
The second line contains string B.
The length of both strings will not be greater than 100.
Output
A single line contains one integer representing the answer.
Sample Input
zzzzzfzzzzz
abcdefedcba
abababababab
cdcdcdcdcdcd
Sample Output
6
7
分析:
先假设A和B所有的对应位置都不相等,然后令dp[l][r]表示l~r这个区间最小操作的次数
然后对一个区间[l ,r] 我们可以考虑,首先将dp[l][r]=dp[l+1][r]+1;表示需要在上一个区间的基础上,额外操作一次。然后枚举一个k
从l+1到r,如果b[k]==b[l] 那么 l这个位置 就可以在刷k的时候 刷上,而且不影响之前的区间。
既 dp[l][r]=min(dp[l][r],dp[l][k]+dp[k+1][r])
然后,我们令ans[i]表示A的第1到第i 刷成和B相同的 需要的最小的操作次数
若a[i]==b[i] 则 ans[i]=ans[i-1] (因为不需要刷)
else 枚举一个j 表示这个ans[i]从哪个答案转移过来
ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
#include<math.h>
using namespace std;
char a[110],b[110];
int dp[110][110];
int ans[110];
int main()
{
while(~scanf("%s",a+1))
{
int n=strlen(a+1);
scanf("%s",b+1);
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
for(int j=i;j<=n;j++)
{
dp[i][j]=j-i+1;
}
for(int j=1;j<=n;j++)
for(int i=j;i>=1;i--)
{
dp[i][j]=dp[i+1][j]+1;
for(int k=i+1;k<=j;k++)
{
if(b[i]==b[k])
{
dp[i][j]=min(dp[i][j],dp[i+1][k]+dp[k+1][j]);
}
}
}
for(int i=1;i<=n;i++)
{
ans[i]=dp[1][i];
if(a[i]==b[i])
{
ans[i]=min(ans[i],ans[i-1]);
}
else
for(int j=1;j<i;j++)
{
ans[i]=min(ans[i],ans[j]+dp[j+1][i]);
}
}
printf("%d\n",ans[n]);
}
return 0;
}