標籤:des style http color java os io strong
Parenthese sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 716 Accepted Submission(s): 335
Problem Descriptionbobo found an ancient string. The string contains only three charaters -- "(", ")" and "?".
bobo would like to replace each "?" with "(" or ")" so that the string is valid (defined as follows). Check if the way of replacement can be uniquely determined.
Note:
An empty string is valid.
If S is valid, (S) is valid.
If U,V are valid, UV is valid.
InputThe input consists of several tests. For each tests:
A string s1s2…sn (1≤n≤106).
OutputFor each tests:
If there is unique valid string, print "Unique". If there are no valid strings at all, print "None". Otherwise, print "Many".
Sample Input
??????(??
Sample Output
UniqueManyNone
AuthorXiaoxu Guo (ftiasch)
Source2014 Multi-University Training Contest 5
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=1001000;char str[maxn];int num[maxn],prefix[maxn],suffix[maxn];int pre0[maxn],suf0[maxn];void init(){ memset(num,0,sizeof(num)); memset(prefix,0,sizeof(prefix)); memset(suffix,0,sizeof(suffix)); memset(pre0,0,sizeof(pre0)); memset(suf0,0,sizeof(suf0));}int main(){while(scanf("%s",str)!=EOF){ int n=strlen(str); if(n%2==1) { puts("None"); continue; } init(); for(int i=0;i<n;i++) { if(str[i]=='(') num[i+1]=1; else if(str[i]==')') num[i+1]=-1; else if(str[i]=='?') num[i+1]=0; } bool flag=true; for(int i=1;i<=n;i++) { if(num[i]) prefix[i]=prefix[i-1]+num[i]; else prefix[i]=prefix[i-1]+1; if(prefix[i]<0) { flag=0; break; } } for(int i=n;i>=1;i--) { if(prefix[i]<=1) pre0[i]=pre0[i+1]+1; else pre0[i]=pre0[i+1]; } if(flag==false) { puts("None"); continue; } for(int i=n;i>=1;i--) { if(num[i]) suffix[i]=suffix[i+1]-num[i]; else suffix[i]=suffix[i+1]+1; if(suffix[i]<0) { flag=false; break; } } for(int i=1;i<=n;i++) { if(suffix[i]<=1) suf0[i]=suf0[i-1]+1; else suf0[i]=suf0[i-1]; } if(flag==false) { puts("None"); continue; } int cnt=0; for(int i=2;i<n;i++) { if(num[i]==0) { if( (prefix[i]>=2&&pre0[i]==0) && (suffix[i]>=2&&suf0[i]==0) ) cnt++; } } if(cnt) puts("Many"); else puts("Unique");}return 0;}