HDU 6170 Two strings (DP)

Posted 2020-10-04 dwtfukgv

题意：给定两个字符串，问你是不是匹配，这不是完全的正则表达式，而且题意有点模糊，‘.‘能匹配任意字符。‘*‘能匹配前面一个字符重复0-无数多次，如果是 . *  这样的是先匹配 .，再匹配*。

析：dp[i][j] 表示 第一个串匹配到 i 第二串匹配到 j，是不是能。

如果是a[i] == b[j] 那就是 dp[i-1][j-1] 

如果 b[j] == ‘.‘ 就是 dp[i-1][j-1]

如果 b[j] == ‘*‘ 那么要分情况，一种是 a[i] == a[i-1]  就是 dp[i-1][j-1] | dp[i-1][j] 

另一种无所谓 就是  dp[i][j-1] | dp[i][j-2]

提供几个数据：

ababb
ababb*

yes

b
ba*

yes

aab
c*a*b

yes

b
baa*

no

abc
.*

no

abbdaaabccc
.*...*.b*.*

no

aac
c*aac

yes

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

aa*aa.*aaa

yes

aaaaaaaaaaaaaaaaaaaaaaaa

a*aa*a*a*a*a*aa*a*a*a*a*a*a*a*

yes

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2500 + 10;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
    return r > 0 && r <= n && c > 0 && c <= m;
}

bool dp[maxn][maxn];
char a[maxn], b[maxn];

int main(){
  freopen("1010.in", "r", stdin);
  freopen("out.txt", "w", stdout);
  int T;  cin >> T;
  while(T--){
    scanf("%s", a+1);
    scanf("%s", b+1);
    ms(dp, 0);
    dp[0][0] = true;
    n = strlen(a+1);
    m = strlen(b+1);
    for(int i = 0; i <= n; ++i){
      for(int j = 1; j <= m; ++j){
        if(a[i] == b[j])  dp[i][j] |= dp[i-1][j-1];
        else if(b[j] == ‘.‘)  dp[i][j] |= dp[i-1][j-1];
        else if(b[j] == ‘*‘ && a[i] == a[i-1])  dp[i][j] |= dp[i-1][j-1] | dp[i-1][j];
        if(b[j] == ‘*‘)  dp[i][j] |= dp[i][j-1] | dp[i][j-2];
      }
    }
    puts(dp[n][m] ? "yes" : "no");
  }
  return 0;
}