标签：lin bool input %s for ble hit div pos
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 nn strings so that the concatenation of the strings represents a valid string.
Validity of strings are defined as follows:
For example, "()()" and "(())" are valid, while "())" and "((()" are not valid.
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 a line with "Yes" (without quotes) if you can make a valid string, or "No" otherwise.
3 ()(()(( ))()()(() )())(())
Yes
2 ))()(( ))((())(
No
1 // 思维+贪心 2 // Aizu - 2681 3 #include <iostream> 4 #include <cstdio> 5 #include <cstring> 6 #include <string> 7 #include <algorithm> 8 #include <utility> 9 #include <vector> 10 #include <map> 11 #include <queue> 12 #include <stack> 13 #include <cstdlib> 14 typedef long long ll; 15 #define lowbit(x) (x&(-x)) 16 #define ls l,m,rt<<1 17 #define rs m+1,r,rt<<1|1 18 using namespace std; 19 const int N=1e2+9; 20 int n; 21 char s[N]; 22 typedef pair<int,int>P; 23 P p[N]; 24 P solve(char s[]) 25 { 26 int l=0,r=0; 27 int i; 28 for(i=0;s[i];i++){ 29 if(s[i]==‘(‘) l++; 30 else if(l) l--; 31 else r++; 32 } 33 return P(l,r);//l为( 数目，r为)数目 34 } 35 bool vis[N]; 36 int main() 37 { 38 scanf("%d",&n); 39 int cnt=0; 40 for(int i=0;i<n;i++) 41 { 42 scanf("%s",s); 43 p[i]=solve(s); 44 if(!p[i].second) cnt+=p[i].first,i--,n--; 45 // cnt 为只有(的字符串的（的和 46 } 47 int ans=0,ret=1; 48 memset(vis,0,sizeof(vis)); 49 int mx,id; 50 for(int i=0;i<n&&ret;i++){ 51 id=-1,mx=-120; 52 for(int j=0;j<n;j++) 53 { 54 if(vis[j]||ans<p[j].first) continue; 55 int k=p[j].second-p[j].first; 56 if(mx<k) mx=k,id=j; 57 } 58 //第一个放在最右边 ，依次往左（前）放。 59 // mx 提供最多的右括号，（贪心） 60 if(id==-1) ret=0; 61 vis[id]=1; 62 ans+=mx;//ans 为这个字符串的净提供右括号数目。 63 } 64 // printf("%d %d %d\n",ret,ans,cnt); 65 if(ret&&ans==cnt) printf("Yes\n"); 66 else printf("No\n"); 67 return 0; 68 }
标签：lin bool input %s for ble hit div pos
原文地址：https://www.cnblogs.com/tingtin/p/9350235.html
