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poj 1458 Common Subsequence
Description
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
Input
The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.
Output
For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.
Sample Input
abcfbc abfcab programming contest abcd mnp
Sample Output
4 2 0
LCS经典题目,重要的是推出递推公式,考虑到是否需要重新计算减少空间,直接贴代码
本题wa了两次,第一次是maxn开大了10010,导致超时,第二次开到110,太小,导致wa
ac之后用的是1010
/* time:11/24 2014 poj:1458 author:cj
accept*/#include<iostream>#include<cstdio>#include<string>#include<cstring>using namespace std;const int maxn=1010;char x[maxn],y[maxn];int dp[maxn][maxn];int m,n;int i,j;int main(){ while(scanf("%s",x+1)!=EOF){ scanf("%s",y+1); memset(dp,0,sizeof(dp)); m=strlen(x+1); n=strlen(y+1); //dp[0][0]=0; /*for(int i=0;i<m;i++){ dp[i][0]=0; }*/ /*for(int j=0;j<n;j++){ //cout<<y[j]<<endl; printf("%c ",y[j]); } /*cout<<x[m]<<y[n]<<endl; if(x[0]==y[0]){ dp[0][0]=1; }*/ for(int i=1;i<=m;i++){ for(int j=1;j<=n;j++){ if(x[i]==y[j]){ dp[i][j]=dp[i-1][j-1]+1; //cout<<dp[i][j]<<endl; } else{ dp[i][j]=max(dp[i-1][j],dp[i][j-1]); } } } //cout<<x<<"***"<<y<<endl;*/ //printf("%s m= %d\n",x,m); //printf("%s n= %d\n",y,n); cout<<dp[m][n]<<endl; } return 0;}
poj 1458 Common Subsequence
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