HDU 6170 Two String 动态规划

题目链接

2017多校训练9-1010

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Problem Description
Giving two strings and you should judge if they are matched.
The first string contains lowercase letters and uppercase letters.
The second string contains lowercase letters, uppercase letters, and special symbols: “.” and “*”.
. can match any letter, and * means the front character can appear any times. For example, “a.b” can match “acb” or “abb”, “a*” can match “a”, “aa” and even empty string. ( “*” will not appear in the front of the string, and there will not be two consecutive “*”.
.

Input
The first line contains an integer T implying the number of test cases. (T≤15)
For each test case, there are two lines implying the two strings (The length of the two strings is less than 2500).


Output
For each test case, print “yes” if the two strings are matched, otherwise print “no”.


Sample Input
3
aa
a*
abb
a.*
abb
aab


Sample Output
yes
yes
no

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可以去找LeetCode Problem 10，题目极其类似

本题.*不能匹配ab。

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思路

考虑dp(i,j)表示s串的前i个字符能否匹配p串的前j个字符。

状态转移方程：

没有出现*
dp[i][j] = dp[i-1][j-1] ,    s[i-1] == p[j-1]或者p[j-1]=='.'
如果出现*
则dp[i][j] = dp[i][j-2]（0次缩放），dp[i][j-1](1次缩放)，dp[i-1][j](多次缩放)
其他情况
dp[i][j] = 0

边界值是一定匹配失败，还有匹配空串的情况

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <string>
#include <math.h>
#include <bitset>
#include <ctype.h>
using namespace std;
typedef pair<int,int> P;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-9;
const int N = 3000 + 5;
const int mod = 1e9 + 7;
int T;
char s1[N],s2[N];
int dp[N][N];
bool ismatch(const char *s, const char *p)
{
    int ls = strlen(s);
    int lp = strlen(p);
    for(int i = 0; i <= ls; i++)
        for(int j = 0; j <= lp; j++)
        dp[i][j] = 1;
    for(int i = 1; i <= ls; i++)
        dp[i][0] = 0;
    for(int j = 1; j <= lp; j++)
    {
        if(p[j-1] == '*') dp[0][j] = dp[0][j-2];
        else  dp[0][j] = 0;
    }

    for(int i = 1; i <= ls; i++)
    {
        for(int j = 1; j <= lp; j++)
        {
            if(p[j-1] == '.' || s[i-1] == p[j-1])
                dp[i][j] = dp[i-1][j-1];
            else if(p[j-1] == '*')
            {
                dp[i][j] = (j >= 2 && dp[i][j-2]) || (dp[i][j-1]) || (dp[i-1][j]&&(s[i-1] == p[j-2] || (p[j-2] == '.' && i >= 2 && s[i-1] == s[i-2])));
            }
            else
                dp[i][j] = 0;
        }
    }
    return dp[ls][lp];
}

int main()
{
    scanf("%d", &T);
    while(T--)
    {
        scanf("%s%s", s1, s2);
        printf("%s
", ismatch(s1,s2) ? "yes" : "no");
    }
    return 0;
}