poj 2192

题意：给出两串，两串顺序不变看能否组成第三个串。

此题深搜和DP都能解决：

深搜的话需要几个强有力剪枝条件

1、第三个串最后一个字符要么是串1的最后一个字符，要么是串2的最后一个字符

2、按照串1的顺序对串3进行搜索，若不匹配则该字符必是串2的下一个字符。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char first[202],second[202],third[402],Left[401];
int sign[402];
bool flag;
int check()
{
    int i,count=0;
    int k=strlen(third);
    for(i=0;i<k;i++)
        if(!sign[i]) Left[count++]=third[i];
    Left[count]='\0';
    if(strcmp(Left,second)==0) return 1;
    return 0;
}
int dfs(int f,int s,int t)
{
    if(f>=strlen(first))
    {
        if(check()) flag=true;
        return 0;
    }
    if(flag) return 0;
    if(first[f]==third[s])
    {
        sign[s]=1;
        if(s<strlen(third)) dfs(f+1,s+1,t);
        sign[s]=0;
    }
    else
    {
        if(third[s]!=second[t]) return 0;//剪枝2
    }
    if(!flag && s<strlen(third)) dfs(f,s+1,t+1);
    return 0;
}
int main()
{
    int len1,len2,len3,Case,count=0;
    scanf("%d",&Case);
    while(Case--)
    {
        count++;
        flag=false;
        scanf("%s %s %s",first,second,third);
        memset(sign,0,sizeof(sign));
 
        len1=strlen(first);
        len2=strlen(second);
        len3=strlen(third);
 
        if(len1+len2!=len3)
        {
            printf("Data set %d: no\n",count);
            continue;
        }
        if(third[len3-1]!=first[len1-1] && third[len3-1]!=second[len2-1])// 剪枝1
        {
            printf("Data set %d: no\n",count);
            continue;
        }
        dfs(0,0,0);
        if(flag) 
            printf("Data set %d: yes\n",count);
        else
            printf("Data set %d: no\n",count);
    }
    return 0;
}

若用DP来作先定义res[i][j]=1表示串1前i个字符和串2的前j个字符能组成串3的前i+j个字符，res[i][j]=0则不能。

状态转移方程如下：

Res[i][j]= (third[i+j]==first[i] && res[i-1][j]==1) ||(third[i+j]==second[j]&&res[i][j-1]==1)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char first[201],second[201],third[401];
int res[201][201];
int init(int n,int m)
{
    int i;
    for(i=1;i<=m;i++)
        if(second[i]==third[i]) res[0][i]=1;
        else break;
    for(i=1;i<=n;i++)
        if(first[i]==third[i]) res[i][0]=1;
        else break;
    return 0;
}
int dp(int n,int m)
{
    int i,j;
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
        {
            if(third[i+j]==first[i] && res[i-1][j]) res[i][j]=1;
            if(third[i+j]==second[j] && res[i][j-1]) res[i][j]=1;
        }
    if(res[n][m]) return 1;
    return 0;
}
int main()
{
    int n,len1,len2,count=0;;
    scanf("%d",&n);
    while(n--)
    {
        count++;
        scanf("%s %s %s",first+1,second+1,third+1);
        len1=strlen(first+1);
        len2=strlen(second+1);
        memset(res,0,sizeof(res));
        init(len1,len2);
    
        if(dp(len1,len2))
            printf("Data set %d: yes\n",count);
        else
            printf("Data set %d: no\n",count);
    }
    return 0;
}