Codeforces Beta Round #5——C. Longest Regular Bracket Sequence
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这道题卡了很久没想出来方法,又看题解了= =
题意:
现在给你一个串,然后让你求出这个串中最长的匹配长度以及是同样是这个最大长度的子串的个数。
思路:
第一种方法:
可以用栈来做,就像判断括号是否匹配那样,当遇到'('时,我们把当前点的坐标给压入栈中,然后遇到')'时,我们从栈中弹出,然后判断是否匹配。
求距离的话就是像尺取法一样根据左右点的取值然后来算出长度。
#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<vector>#include<map>#include<stack>using namespace std;#define maxn 1000010stack<int> sk;char a[maxn];int l[maxn];int main(){scanf("%s",a);int len=strlen(a);int ans=-1,num=0;for(int i=0;i<len;i++){if(a[i]=='('){sk.push(i);}else{if(!sk.empty()){int k=sk.top();sk.pop();if(k>0) l[i]=l[k-1];//注意这里还有一种特殊情况就是我们必须把前面的长度也得累加起来,这样才能保证是最大的; else l[i]=0;l[i]+=i-k+1;if(l[i]>ans){ans=l[i];num=1;}else if(ans==l[i]) num++;}}}if(ans==-1) printf("0 1\n");else printf("%d %d\n",ans,num);}/*)((())))(()())*/
第二种方法我想应该是dp,但是我现在不想看题解,希望能够有自己独立思考的时间,所以等我想出来了解法,我会补上这种做法的。加油!
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