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https://github.com/Kiritow/OJ-Problems-Source.git
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37 lines
1.2 KiB
C++
37 lines
1.2 KiB
C++
#include <cstdio>
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#include <cstring>
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#include <algorithm>
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using namespace std;
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int n,m,al,bl,cl,lim;
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char a[205],b[205],c[405];
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bool dp[205][205];//dp[i][j]表示串a的前i个字符与串b的前j个字符是否能形成串c的前i+j个字符
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int main() {
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int T,kase=0;
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scanf("%d",&T);
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while(kase<T) {
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scanf("%s%s%s",a+1,b+1,c+1);
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al=strlen(a+1);
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bl=strlen(b+1);
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cl=al+bl;
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memset(dp,false,sizeof(dp));
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dp[0][0]=true;
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for(int k=1;k<=cl;++k) {//枚举串c的字符
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lim=min(k,al);
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for(int i=max(1,k-bl);i<=lim;++i) {//在a串上进行状态转移
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if(dp[i-1][k-i]&&a[i]==c[k])//如果串a前i-1个字符与串b前k-i个字符能构成串c的前k个字符，并且当前a[i]==c[k]
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dp[i][k-i]=true;
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}
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lim=min(k,bl);
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for(int j=max(1,k-al);j<=lim;++j) {//在b串上进行状态转移
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if(dp[k-j][j-1]&&b[j]==c[k])//如果串a前k-j个字符与串b前j-1个字符能构成串c的前k个字符，并且当前b[j]==c[k]
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dp[k-j][j]=true;
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}
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}
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printf("Data set %d: %s\n",++kase,dp[al][bl]?"yes":"no");
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}
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return 0;
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}
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