1010. Zipper

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
Copy sample input to clipboard
3
cat tree tcraete
cat tree catrtee
cat tree cttaree
Sample Output
Data set 1: yes
Data set 2: yes
Data set 3: no

CPP

#include <iostream>
#include <string>
#include <cstring>
using namespace std;

const int N = 205;

string a, b, c;
bool dp[N][N];

int main()
{
    int cases;
    int num = 0;

    cin >> cases;

    while (cases--)
    {
        num++;
        cin >> a >> b >> c;

        memset(dp, false, sizeof(dp));

        dp[0][0] = true;

        for (int i = 0; i < b.size(); i++)
            if (b[i] == c[i])
                dp[0][i+1] = true;//
            else
                break;

        for (int i = 0; i < a.size(); i++)
            if (a[i] == c[i])
                dp[i+1][0] = true;
            else
                break;
        //确定c最初几个字母的归属
        for (int i = 1; i <= a.size(); i++)
            for (int j = 1; j <= b.size(); j++)
                dp[i][j] = (dp[i-1][j] && c[i+j-1] == a[i-1]) || (dp[i][j-1] && c[i+j-1] == b[j-1]);
        //通过二维数组来确定
        //判断条件需学会！
        if (dp[a.size()][b.size()])
            cout << "Data set "<< num << ": yes" << endl;
        else
            cout << "Data set "<< num << ": no" << endl;
    }
    return 0;
}

C （有未知bug！）

#include <stdio.h>
#include <string.h>
int main() {
    int n;
    scanf("%d", &n);
    int k;
    for (k = 1; k <= n; k++) {
        char a[200], b[200], c[200];
        scanf("%s%s%s", a, b, c);
        int dp[200][200];
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        for (int i = 0; i < strlen(b); i++) {
            if (b[i] == c[i]) {
                dp[0][i + 1] = 1;
            } else {
                break;
            }
        }
        for (int i = 0; i < strlen(a); i++) {
            if (a[i] == c[i]) {
                dp[i + 1][0] = 1;
            } else {
                break;
            }
        }
        for (int i = 1;i <= strlen(a); i++) {
            for (int j = 1; j <= strlen(b); j++) {
                dp[i][j] = (dp[i - 1][j] && c[i + j - 1] == a[i - 1]) || (dp[i][j - 1] && c[i + j - 1] == b[j - 1]);
            }
        }
        if (dp[strlen(a)][strlen(b)]) {
            printf("Data set %d: yes\n", k);
        } else {
            printf("Data set %d: no\n", k);
        }
    }
    return 0;
}

此题是通过定义一个二维数组，前一个变量确定a后一个变量确定b
判断条件尤其重要！！