FOJ 1381 Regular Expressions

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1010;
const int inf  = 0x3f3f3f3f;
char s1[maxn], s2[maxn];
int dp[maxn][maxn];

// 题目链接：http://acm.fzu.edu.cn/problem.php?pid=1381

/*
    dp[i][j]: 代表第一个regex的前i个字符和第二个字符串的前j个字符匹配的最小长度
*/

int main() {

    //freopen("aa.in", "r", stdin);
    int l1, l2;
    while(scanf("%s %s", s1 + 1, s2 + 1) != EOF) {
        l1 = strlen(s1 + 1);
        l2 = strlen(s2 + 1);
        dp[0][0] = 0;
        for(int i = 1; i <= l2; ++i) {
            if(s2[i] >= 'A' && s2[i] <= 'Z') {
                dp[0][i] = inf;
            } else if(s2[i] == '*') {
                dp[0][i] = dp[0][i-1];
            } else {
                dp[0][i] = inf;
            }
        }
        for(int i = 1; i <= l1; ++i) {
            if(s1[i] >= 'A' && s1[i] <= 'Z') {
                dp[i][0] = inf;
            } else if(s1[i] == '*') {
                dp[i][0] = dp[i-1][0];
            } else {
                dp[i][0] = inf;
            }
        }
        for(int i = 1; i <= l1; ++i) {
            for(int j = 1; j <= l2; ++j) {
                if(s1[i] == '*' && s2[j] == '*') {
                    dp[i][j] = min(dp[i-1][j-1], min(dp[i][j-1], dp[i-1][j]));
                } else if(s1[i] == '*'){
                    dp[i][j] = min(dp[i-1][j], min(dp[i][j-1] + 1, dp[i-1][j-1] + 1));
                } else if(s2[j] == '*') {
                    dp[i][j] = min(dp[i][j-1], min(dp[i-1][j] + 1, dp[i-1][j-1] + 1));
                } else if(s1[i] == '?' || s2[j] == '?'){
                    dp[i][j] = dp[i-1][j-1] + 1;
                } else {
                    if(s1[i] == s2[j]) {
                        dp[i][j] = dp[i-1][j-1] + 1;
                    } else {
                        dp[i][j] = inf;
                    }
                }
            }
        }
        if(dp[l1][l2] >= inf) {
            printf("No Solution!\n");
        } else {
            printf("%d\n", dp[l1][l2]);
        }
    }
    return 0;
}