pku 1458 Common Subsecquence(DP)

 

题目大意：给定两个字符串，求其最大的公共子串。在这里子串的定义是：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 >.

分析：简单DP

if(first[i]=second[j]) //在最大公共子串中，first[i]和second[j]恰好配对

count[i][j]=count[i-1][j-1]+1;

else count[i][j]=max(count[i-1][j],count[i][j-1]); //否则

 

#include <iostream> using namespace std; #define Max(x,y) (x<y?y:x) short count[1005][1005],first_n,second_n,k; char first[1005],second[1005]; void DP(); int main() { while(scanf("%s",first+1)!=EOF) { scanf("%s",second+1); first_n=strlen(first+1); second_n=strlen(second+1); DP(); printf("%hd/n",count[first_n][second_n]); } return 0; } void DP() { memset(count,0,sizeof(count)); for(int i=1;i<=first_n;i++) for(int j=1;j<=second_n;j++) if(first[i]==second[j]) count[i][j]=count[i-1][j-1]+1; else count[i][j]=Max(count[i][j-1],count[i-1][j]); }