hdu4915 Parenthese sequence 2014 Multi-University Training Contest 5

Parenthese sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 28    Accepted Submission(s): 5

Problem Description

bobo 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.

 

Input

The input consists of several tests. For each tests:

A string s₁s₂…s_n (1≤n≤10⁶).

 

Output

For 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

 

Source

2014 Multi-University Training Contest 5

 

Recommend

hujie

 

比较麻烦的题，考虑多种情况，首先排除不行的，然后判断是否能有多个解，比赛的时候有个地方写错wa了5次···

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int N=1000005;int f1[N],f2[N];char s[N];int main(){    while(scanf("%s",s)!=EOF){        int l=strlen(s);        if(l%2){            printf("None\n");            continue;        }        if(s[0]==')'||s[l-1]=='('){            printf("None\n");            continue;        }        int flag=0;        memset(f1,0,sizeof(f1));        for(int i=1;i<l;i++){            if(s[i]==')')f1[i]++;            f1[i]+=f1[i-1];            if(f1[i]>(i+1)/2){                printf("None\n");                flag=1;                break;            }        }        if(flag)continue;        memset(f2,0,sizeof(f2));        for(int i=l-2;i>=0;i--){            if(s[i]=='(')f2[i]++;            f2[i]+=f2[i+1];            if((l-i)/2<f2[i]){            printf("None\n");            flag=1;            break;            }        }        if(flag)continue;        int vis1=0,vis2=0;        for(int i=l-1;i>=0;i--){            if(f1[i]==(i+1)/2)break;            if(s[i]=='?')vis1=i;        }        for(int i=0;i<l;i++){            if(f2[i]==(l-i)/2)break;            if(s[i]=='?')vis2=i;        }        if(vis1&&vis2&&vis2>vis1) printf("Many\n");        else printf("Unique\n");    }    return 0;}