​https://ac.nowcoder.com/acm/problem/21303​

 

 

 

n^3正解​​http://www.perfectpan.org/archives/2520​

 

自己n^5的傻逼代码

#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1e2+10;

int dp[maxn][maxn][maxn];
int pre[maxn];
char ch1[maxn],ch2[maxn];
int n1,n2;

bool judge(int p,int l,int r)
{
int i;
for(i=l;i<=r;i++){
if(p+pre[i]-pre[l-1]<0) return 0;
}
return 1;
}

int main()
{
int i,j,k,l,flag;
scanf("%s%s",ch1+1,ch2+1);
n1=strlen(ch1+1),n2=strlen(ch2+1);
for(i=1;i<=n1;i++){
pre[i]=pre[i-1];
if(ch1[i]=='(') pre[i]++;
else pre[i]--;
}
memset(dp,0,sizeof(dp));
dp[0][0][0]=1;
for(i=1;i<=n2;i++){//dp[i-1][k][] -> dp[i][j][]
for(j=i;j+(n2-i)<=n1;j++){
if(ch2[i]==ch1[j]){
for(k=0;k<=j-1;k++){
if(ch2[i-1]==ch1[k]){
for(l=0;l<=k;l++){
//if(judge(l,k+1,j-1)){
//if(l+pre[j-1]-pre[k]>=0){
if(judge(l,k+1,j-1)&&l+pre[j-1]-pre[k]==0){
dp[i][j][l+pre[j-1]-pre[k]]|=dp[i-1][k][l];
//if(dp[i-1][k][l]) printf("*%d %d %d %d %d %d*\n",i,j,l+pre[j-1]-pre[k],i-1,k,l);
}
}
}
}
}
}
}
flag=0;
for(j=1;j<=n1;j++){
flag|=dp[n2][j][0];
}
if(flag) printf("Possible\n");
else printf("Impossible\n");
return 0;
}