题意:
给你两个字符串,求一个最短的串,使得这两个串是目标串的子串。
思路:
这题可以借鉴LCS的思想,对于求最小串的长度,就很简单了
1. s1[i] == s2[j] l[i, j] = l[i-1, j-1];
2. s1[i] != s2[j] l[i, j] = min(l[i-1, j], l[i, j-1]) + 1;对于求有多少种组合方法,定义dp[i, j]表示两个串的前i, j个有多少个组合方法。
1. s1[i] == s2[j] 显然s1[i],s2[j]在最后的摆放次序已经无关了
dp[i, j] = dp[i-1, j-1];
2. s1[i] != s2[j] 此时要根据l[i, j-1]和l[i-1, j]的长度进行判断
(1). l[i-1][j] < l[i][j-1] dp[i][j] = dp[i-1][j];
(2). l[i-1][j] > l[i][j-1] dp[i][j] = dp[i][j-1];
(3). l[i-1][j] == l[i][j-1] dp[i][j] = (dp[i-1][j] + dp[i][j-1]);
AC代码
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 40;
char str1[N],str2[N];
int l[N][N],dp[N][N];
int main() {
int T ,cas = 1;
scanf("%d", &T); getchar();
while(T--) {
gets(str1);
gets(str2);
int len1 = strlen(str1), len2 = strlen(str2);
memset(dp,0,sizeof(dp));
memset(l,0,sizeof(l));
for(int i = 1; i < N; i++) {
l[i][0] = l[0][i] = i;
dp[i][0] = dp[0][i] = 1;
}
dp[0][0] = 1;
for(int i = 1; i <= len1; i++) {
for(int j = 1; j <= len2; j++) {
if(str1[i-1] == str2[j-1]) {
l[i][j] = l[i-1][j-1] + 1;
dp[i][j] = dp[i-1][j-1];
}else {
l[i][j] = min(l[i-1][j],l[i][j-1]) + 1;
if(l[i-1][j] < l[i][j-1]) {
dp[i][j] = dp[i-1][j];
}else if(l[i-1][j] > l[i][j-1]) {
dp[i][j] = dp[i][j-1];
}else {
dp[i][j] = (dp[i-1][j] + dp[i][j-1]);
}
}
}
}
printf("Case #%d: %d %d\n",cas++,l[len1][len2],dp[len1][len2]);
}
return 0;
}