Description
E - Parentheses
Problem Statement
You are given nn strings str1,str2,…,strnstr1,str2,…,strn, each consisting of ( and ). The objective is to determine whether it is possible to permute the nnstrings so that the concatenation of the strings represents a valid string.
Validity of strings are defined as follows:
- The empty string is valid.
- If AA and BB are valid, then the concatenation of AA and BB is valid.
- If AA is valid, then the string obtained by putting AA in a pair of matching parentheses is valid.
- Any other string is not valid.
For example, "()()" and "(())" are valid, while "())" and "((()" are not valid.
Input
The first line of the input contains an integer nn (1≤n≤1001≤n≤100), representing the number of strings. Then nn lines follow, each of which contains stristri (1≤∣stri∣≤1001≤∣stri∣≤100). All characters in stristri are ( or ).
Output
Output a line with "Yes" (without quotes) if you can make a valid string, or "No" otherwise.
Sample Input 1
3
()(()((
))()()(()
)())(())
Output for the Sample Input 1
Yes
Sample Input 2
2
))()((
))((())(
Output for the Sample Input 2
No
题意：每出现一对()时可以消去，现在给你n个由（和）组成的字符串，问能否通过安排这个n个字符串的位置使得
最后没有任何字符留下；
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int inf =0x7f7f7f7f;
const double pi=acos(-1);
const int maxn=40000;
struct node {
int x,y;
bool operator<(const node&a) const{
return this->x<a.x;
}
};
vector<node> A,B;
char s[10000];
int main()
{
int n;
while(~scanf("%d",&n))
{
A.clear();B.clear();
for(int i=1;i<=n;i++)
{
scanf("%s",s);
int l=0,r=0;
for(int j=0;s[j]!=' ';j++)
{
if(s[j]=='(') r++;
else if(!r) l++;
else r--;
}
if(l>r) B.push_back(node{r,l-r});
else A.push_back((node){l,r-l});
}
sort(A.begin(),A.end());
sort(B.begin(),B.end());
int cntA=0,cntB=0;
bool flag=true;
for(int i=0;i<A.size();i++)
{
if(A[i].x>cntA) {flag=false;break;}
cntA+=A[i].y;
}
for(int i=0;i<B.size();i++)
{
if(B[i].x>cntB) {flag=false;break;}
cntB+=B[i].y;
}
if(cntA!=cntB) flag=false;
printf("%s
",flag?"Yes":"No");
}
return 0;
}
分析：这道题做了很久，其实我的思路大部分是正确的，有几个值得改进的地方：
1.字符串的处理上，我是直接对字符进行移动，而简洁的做法就是直接统计；
我的
while(1)
{
int flag=0;
for(int i=0;s[i+1]!=' ';i++)
{
if(s[i]=='('&&s[i+1]==')')
{
int j;
for(j=i+2;s[j]!=' ';j++)
s[j-2]=s[j];
s[j-2]=' ';
flag=1;
}
if(flag) break;
}
if(!flag) return;
}
简洁：
for(int j=0;s[j]!=' ';j++)
{
if(s[j]=='(') r++;
else if(!r) l++;
else r--;
}
2.最后对右边的处理也不是任意的，比如(((((和右边的)))( ，))))((( ， ))左边的与右边的不能任意结合
我的wa代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int inf =0x7f7f7f7f;
const double pi=acos(-1);
const int maxn=40000;
char s[100000+10];
int cntl[14],cntr[14],n,l=0,r=0,kk,p,flag[15],used[15];
void init()
{
while(1)
{
int flag=0;
for(int i=0;s[i+1]!=' ';i++)
{
if(s[i]=='('&&s[i+1]==')')
{
int j;
for(j=i+2;s[j]!=' ';j++)
s[j-2]=s[j];
s[j-2]=' ';
flag=1;
}
if(flag) break;
}
if(!flag) return;
}
}
struct node{
int id,cntr;
};
bool operator<(node a,node b)
{
return a.cntr<b.cntr;
}
void solve()
{
int l=0;
for(int i=1;i<=n;i++)
if(cntl[i]&&!cntr[i]) {l+=cntl[i];used[i]=1;}
if(l==0) {printf("No
");return;}
priority_queue<node> q;
for(int i=1;i<=n;i++)
if(!used[i]&&cntl[i]>=cntr[i])
q.push((node){i,cntr[i]});
while(q.size())
{
node u=q.top();q.pop();
int i=u.id;
if(l<cntr[i]) {printf("No
");return;}
l+=cntl[i];
l-=cntr[i];
used[i]=1;
}
for(int i=1;i<=n;i++)
if(!used[i]&&cntl[i])
{
if(l<cntr[i]) {printf("No
");return;}
l-=cntr[i];
l+=cntl[i];
used[i]=1;
}
for(int i=1;i<=n;i++)
if(!used[i])
{
if(l<cntr[i]) {printf("No
");return;}
l-=cntr[i];
used[i]=1;
}
if(l) {printf("No
");return;}
else {printf("Yes
");return;}
}
int main()
{
while(~scanf("%d",&n))
{
MM(cntl,0);MM(cntr,0);MM(used,0);
p=n;l=r=kk=0;
for(int k=1;k<=n;k++)
{
scanf("%s",s);
init();
for(int i=0;s[i]!=' ';i++)
if(s[i]=='(') cntl[k]++;
else if(s[i]==')') cntr[k]++;
if(!cntl[k]&&!cntr[k]) p--;
}
if(p==0) {printf("Yes
");continue;}
solve();
}
return 0;
}