标签：dp
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10225 | Accepted: 3949 |
Description
Let x and y be two strings over some finite alphabet A. We would like to transform x into y allowing only operations given below:
Certainly, we would like to minimize the number of all possible operations.
IllustrationA G T A A G T * A G G C | | | | | | | A G T * C * T G A C G CDeletion: * in the bottom line
Insertion: * in the top line
Change: when the letters at the top and bottom are distinct
This tells us that to transform x = AGTCTGACGC into y = AGTAAGTAGGC we would be required to perform 5 operations (2 changes, 2 deletions and 1 insertion). If we want to minimize the number operations, we should do it like
A G T A A G T A G G C | | | | | | | A G T C T G * A C G C
and 4 moves would be required (3 changes and 1 deletion).
In this problem we would always consider strings x and y to be fixed, such that the number of letters in x is m and the number of letters in y is n where n ≥ m.
Assign 1 as the cost of an operation performed. Otherwise, assign 0 if there is no operation performed.
Write a program that would minimize the number of possible operations to transform any string x into a string y.
Input
The input consists of the strings x and y prefixed by their respective lengths, which are within 1000.
Output
An integer representing the minimum number of possible operations to transform any string x into a string y.
Sample Input
10 AGTCTGACGC 11 AGTAAGTAGGC
Sample Output
4
类似于编辑距离问题。。dp[i][j]代表 x串从1-i与y串从1-j已经匹配好的最少操作数。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#define ll long long
using namespace std;
const int INF = 0x3f3f3f3f;
int dp[1025][1025];
char a[1025],b[1025];
int My_min(int x,int y,int z)
{
return min(min(x,y),z);
}
int main()
{
int lena,lenb;
while(scanf("%d %s %d %s",&lena,a,&lenb,b)!=EOF){
for(int i=0;i<=lenb;i++)
dp[0][i]=i;
for(int i=0;i<=lena;i++)
dp[i][0]=i;
for(int i=1;i<=lena;i++)
for(int j=1;j<=lenb;j++)
dp[i][j]=My_min(dp[i-1][j]+1,dp[i][j-1]+1,a[i-1]==b[j-1]?dp[i-1][j-1]:dp[i-1][j-1]+1);
printf("%d\n",dp[lena][lenb]);
}
return 0;
}标签：dp
原文地址：http://blog.csdn.net/qq_16255321/article/details/40783833
