POJ-1458 Common Subsequence 【LCS模板】

A - Common Subsequence

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu
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Status
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, x ij = 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

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1. 题意：求两个串的最长公共子串；
2. 思路：dp[i][j]代表以a[i] b[j]结尾的最大公共子串；
3. 失误：清0差点忘了，多次测试，定义全局也不行；
4. 代码如下：

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#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=1e3+10;char a[MAXN],b[MAXN];int dp[MAXN][MAXN]; int main(){      int i,j,la,lb;     while(~scanf("%s %s",a+1,b+1)) { //从1开始         memset(dp,0,sizeof(dp));  //初始化         for(i=1;a[i]!='\0';++i){            for(j=1;b[j]!='\0';++j){                if(a[i]==b[j]) dp[i][j]=dp[i-1][j-1]+1;//求LCS                 else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);                lb=j;              }            la=i;//记录长度         }        printf("%d\n",dp[la][lb]);    }     return 0; }