POJ 2192 Zipper简单dp

Zipper

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17952		Accepted: 6384

Description

Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in its original order.

For example, consider forming "tcraete" from "cat" and "tree":

String A: cat
String B: tree
String C: tcraete

As you can see, we can form the third string by alternating characters from the two strings. As a second example, consider forming "catrtee" from "cat" and "tree":

String A: cat
String B: tree
String C: catrtee

Finally, notice that it is impossible to form "cttaree" from "cat" and "tree".

Input

The first line of input contains a single positive integer from 1 through 1000. It represents the number of data sets to follow. The processing for each data set is identical. The data sets appear on the following lines, one data set per line.

For each data set, the line of input consists of three strings, separated by a single space. All strings are composed of upper and lower case letters only. The length of the third string is always the sum of the lengths of the first two strings. The first two strings will have lengths between 1 and 200 characters, inclusive.

Output

For each data set, print:

Data set n: yes

if the third string can be formed from the first two, or

Data set n: no

if it cannot. Of course n should be replaced by the data set number. See the sample output below for an example.

Sample Input

3
cat tree tcraete
cat tree catrtee
cat tree cttaree

Sample Output

Data set 1: yes
Data set 2: yes
Data set 3: no

题意： 给你两个字符串，问你能不能按照每个字符串的顺序拼出第三个来。 首先dfs是可以得出正确答案的，不过得算到明年。所以dp，思路也不难想。

状态只与下标和下标对应的字符有关系，二维就能满足需求了。dp[i][j] means 用s1的前i个字母和s2的前j个字幕能不能凑出s[i + j]能是1， 不能是0。

初状态就是dp[0][0] = 1;

状态转移也好想.

if(s[i + j] == s1[i]) dp[i][j] = dp[i][j] | dp[i - 1][j];

if(s[i + j ] == s2[j]) dp[i][j] = dp[i][j] | dp[i][j - 1];

代码：

/*
 * ac.cpp
 *
 *  Created on: 2016年7月20日
 *      Author: triose
 */


//#include<bits/stdc++.h>
#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<time.h>
#include<map>
#include<set>
using namespace std;
//#define ONLINE_JUDGE
#define eps 1e-8
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define INFL 0x3f3f3f3f3f3f3f3fLL
#define enter putchar(10)
#define rep(i,a,b) for(int i = (a); i < (b); ++i)
#define repe(i,a,b) for(int i = (a); i <= (b); ++i)
#define mem(a,b) (memset((a),b,sizeof(a)))
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define sfs(a) scanf("%s",a)
#define pf(a) printf("%d\n",a)
#define pfd(a,b) printf("%d %d\n",a,b)
#define pfP(a) printf("%d %d\n",a.fi,a.se)
#define pfs(a) printf("%s\n",a)
#define pfI(a) printf("%I64d\n",a)
#define ds(t) int t; sf(t)
#define PR(a,b) pair<a,b>
#define fi first
#define se second
#define LL long long
#define DB double
const double PI = acos(-1.0);
const double E = exp(1.0);
template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<class T> inline T Min(T a, T b) { return a < b ? a : b; }
template<class T> inline T Max(T a, T b) { return a > b ? a : b; }
int n, m;
#define N 1010
char s1[N], s2[N], s[N];
int len1, len2, len;
int dp[N][N];
int main() {
#ifndef ONLINE_JUDGE
	freopen("in.txt","r",stdin);
//	freopen("Out.txt", "w", stdout);
#endif
	ds(t);
	repe(l, 1, t) {
		sfs(s1 + 1); sfs(s2 + 1); sfs(s + 1);
		len1 = strlen(s1 + 1); len2 = strlen(s2 + 1);
		len = strlen(s + 1);
		mem(dp, 0);
		dp[0][0] = 1;
		printf("Data set %d: ", l);
		repe(i, 0, len1) {
			repe(j, 0, len2) {
				if(!i && !j) continue;
				if(s[i + j] == s1[i]) dp[i][j] |= dp[i - 1][j];
				if(s[i + j] == s2[j]) dp[i][j] |= dp[i][j - 1];
			}
		}
		pfs((dp[len1][len2] ? "yes" : "no"));
	}
	return 0;
}