删括号

原题链接:https://ac.nowcoder.com/acm/problem/21303

解题思路:用dp[i][j][k]数组,i表示s1前i个字符,j表示s2的前j个字符,k=s1删去的’(‘-s1删去的’)’;
如果s1前i个字符删去k个 ‘(‘ 与s2相符,则令dp[i][j][k]==true,最终只要满足dp[len1-1][len2-1][0]==true,即为Possible

Code:

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#include<cstring>

using namespace std;
const int N = 200;
int dp[N][N][N];

int (){
string ss,tt;
cin>>ss>>tt;
memset(dp,0,sizeof(dp));
int len1=ss.length();
int len2=tt.length();
dp[0][0][0]=1;
for(int i=0;i<len1;i++){
for(int j=0;j<len2;j++){
for(int k=0;k<len1/2;k++){
if(dp[i][j][k]){

if(k==0&&ss[i+1]==tt[j+1]) dp[i+1][j+1][k]=1;
//下一字符为"(",差值k+1
if(ss[i+1]=='(') dp[i+1][j][k+1]=1;
//凑够(),删去),差值k-1
else if(k) dp[i+1][j][k-1]=1;
}
}
}
}
if(dp[len1-1][len2-1][0]) printf("Possiblen");
else printf("Impossiblen");

return 0;
}