CF5 C. Longest Regular Bracket Sequence
阿新 • • 發佈:2021-07-01
題目傳送門:https://codeforces.com/problemset/problem/5/C
題目大意:
給定一串括號序列,求最長合法括號序列的長度及出現次數
考慮將 '(' 變成 1,將 ')' 變成 -1,故合法括號序列的累加和必然為0
考慮求其字首和,由於括號序列是一一匹配的,故我們可以用vector記錄字首和為 \(k\) 的一系列位置,匹配完後將 '(' 的位置刪除即可
/*program from Wolfycz*/ #include<map> #include<cmath> #include<cstdio> #include<vector> #include<cstring> #include<iostream> #include<algorithm> #define Fi first #define Se second #define ll_inf 1e18 #define MK make_pair #define sqr(x) ((x)*(x)) #define pii pair<int,int> #define int_inf 0x7f7f7f7f using namespace std; typedef long long ll; typedef unsigned int ui; typedef unsigned long long ull; inline char gc(){ static char buf[1000000],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++; } template<typename T>inline T frd(T x){ int f=1; char ch=gc(); for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } template<typename T>inline T read(T x){ int f=1; char ch=getchar(); for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1; for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0'; return x*f; } inline void print(int x){ if (x<0) putchar('-'),x=-x; if (x>9) print(x/10); putchar(x%10+'0'); } const int N=1e6; char s[N+10]; vector<int>pos[N+10]; void init(){for (int i=0;!pos[i].empty();i++) pos[i].clear();} int main(){ // freopen(".in","r",stdin); // freopen(".out","w",stdout); scanf("%s",s+1); int n=strlen(s+1),sum=0; int Max=0,Cnt=0; pos[0].push_back(0); for (int i=1;i<=n;i++){ sum+=s[i]=='('?1:-1; if (sum<0){ init(); sum=0; pos[sum].push_back(i); continue; } if (pos[sum].empty()) pos[sum].push_back(i); else{ int len=i-pos[sum].back(); if (len>Max) Max=len,Cnt=0; Cnt+=(len==Max); pos[sum+1].clear(); } } printf("%d %d\n",Max,!Max?1:Cnt); return 0; }