poj 3356 AGTC
2011-07-24 10:35
239 查看
DescriptionLet x and y be two strings over some finite alphabet A. We would like to transform x into y allowing only operations given below:
Deletion: a letter in x is missing in y at a corresponding position.
Insertion: a letter in y is missing in x at a corresponding position.
Change: letters at corresponding positions are distinct
Certainly, we would like to minimize the number of all possible operations.
Illustration
Insertion: * in the top line
Change: when the letters at the top and bottom are distinctThis 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
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.
InputThe input consists of the strings x and y prefixed by their respective lengths, which are within 1000.
OutputAn integer representing the minimum number of possible operations to transform any string x into a string y.
Sample Input
又解决一道当年遗留下来的题……
这题类似最长公共子串,只是稍微变了一下罢了……注意初始化……
直接上代码,没什么好说的了……多校第一场貌似也有一道这种变题,这两天解决去……
Deletion: a letter in x is missing in y at a corresponding position.
Insertion: a letter in y is missing in x at a corresponding position.
Change: letters at corresponding positions are distinct
Certainly, we would like to minimize the number of all possible operations.
Illustration
A 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 distinctThis 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 Cand 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.
InputThe input consists of the strings x and y prefixed by their respective lengths, which are within 1000.
OutputAn integer representing the minimum number of possible operations to transform any string x into a string y.
Sample Input
10 AGTCTGACGC 11 AGTAAGTAGGCSample Output
4SourceManila 2006
又解决一道当年遗留下来的题……
这题类似最长公共子串,只是稍微变了一下罢了……注意初始化……
直接上代码,没什么好说的了……多校第一场貌似也有一道这种变题,这两天解决去……
#include <stdio.h> #include <algorithm> using namespace std; #define INF 999999999 char str1[1005]; char str2[1005]; int dp[1005][1005]; int main() { int i,j,n,m,x,y; while(scanf("%d",&n)!=EOF) { scanf("%s",str1); scanf("%d",&m); scanf("%s",str2); for (i=0;i<=n;i++) { dp[i][0]=i; } for (i=0;i<=m;i++) { dp[0][i]=i; } dp[0][0]=0; for (i=1;i<=n;i++) { for (j=1;j<=m;j++) { x=i-1; y=j-1; if (str1[x]==str2[y]) dp[i][j]=min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1])+1); else dp[i][j]=min(min(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1])+1; } } printf("%d\n",dp [m]); } return 0; }
相关文章推荐
- POJ 3356 AGTC(DP-最小编辑距离)
- poj 3356 AGTC
- POJ-3356-AGTC
- POJ 3356 AGTC (最小编辑距离)
- POJ-3356 AGTC (最短编辑距离问题)
- POJ 3356 AGTC 最短编辑距离 DP
- POJ 3356 AGTC(最长公共子)
- POJ 3356.AGTC(DP)
- poj 3356 AGTC
- poj 3356 AGTC 简单dp
- POJ 3356 AGTC(最小编辑距离)
- POJ_3356_AGTC
- POJ 3356 AGTC 最短编辑距离
- poj 3356 AGTC(dp 求最短编辑距离)
- POJ 3356 AGTC(算法导论15-5编辑距离) 经典dp
- poj 3356 AGTC(dp,最小编辑距离)
- POJ 3356 AGTC 最短编辑距离
- POJ 3356 AGTC (最长公共子序列)
- POJ 3356 AGTC(经典DP Edit Distance)
- POJ 3356 AGTC(最长公共子序列)