HDU 6170(Two strings-DP)

最新推荐文章于 2019-10-31 23:24:41 发布

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本文介绍了一种用于判断两个字符串是否匹配的算法，其中包括了特殊符号“.”和“*”的使用规则。通过动态规划的方法实现了对给定模式的匹配过程，并提供了一个具体的实现示例。该算法能够处理长度小于2500的字符串，适用于多种应用场景。

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

Source
2017 Multi-University Training Contest - Team 9

做法：直接dp

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<vector>
#include<set>
#include<string>
#include<queue>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
#define MAXN (3000)
bool f[MAXN][MAXN];
char a[MAXN],b[MAXN]; 
int main()
{
//  freopen("J.in","r",stdin);
//  freopen(".out","w",stdout);
    int T=read();
    while(T--) {
        scanf("%s%s",a+1,b+1);
        MEM(f)
        f[0][0]=1;
        int n=strlen(a+1),m=strlen(b+1);
        Rep(i,n+1) For(j,m) {
            f[i][j]=0;
            if (b[j]=='*') f[i][j]|=f[i][j-2];
            if (b[j]=='*') f[i][j]|=f[i][j-1];
            if (b[j]=='.') f[i][j]|=f[i-1][j-1];
            if (!i) continue;
            f[i][j]|=f[i-1][j-1]&(a[i]==b[j]);
            if (b[j]=='*') {
                if (b[j-1]!='.') {
                    if (b[j-1]==a[i]) f[i][j]|=f[i-1][j];
                } else {
                    if (a[i]==a[i-1]) f[i][j]|=f[i-1][j];
                }
            }
//          cout<<f[i][j];
        }
        puts(f[n][m]?"yes":"no");
    }

    return 0;
}