題解

斷環為鏈，把鏈復制兩份
用set維護一下全是0的區間，然後查找x + n / 2附近的區間，附近各一個過不去，最後棄療了改為查附近的兩個，然後過掉了= =

熟練掌握stl的應用，你值得擁有（霧

代碼

#include <bits/stdc++.h>
//#define ivorysi
#define enter putchar(‘\n‘)
#define space putchar(‘ ‘)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eps 1e-8
#define mo 974711
 

#define MAXN 200005
#define pii pair<int,int>
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < ‘0‘ || c > ‘9‘) {
    if(c == ‘-‘) f = -1;
    c = getchar();
    }
    while(c >= ‘0‘
 
 && c <= ‘9‘) {
    res = res * 10 + c - ‘0‘;
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar(‘-‘);x = -x;}
    if(x >= 10) {
    out(x / 10);
    }
    putchar(‘0‘ + x % 10);
}
int N,M;
int A[MAXN * 2],B[MAXN * 2];
char ch[MAXN * 2][3
 
];
set<pii > S;
void Change(int p,int v) {
    if(B[p] == v) return;
    if(B[p] != 0 && v != 0) {B[p] = v;return;}
    if(B[p] == 0) {
    auto k = S.upper_bound(mp(p,4 * N));
    --k;
    pii t = *k;
    S.erase(k);
    if(p - 1 >= t.fi) S.insert(mp(t.fi,p - 1));
    if(p + 1 <= t.se) S.insert(mp(p + 1,t.se));
    }
    else {
    auto k = S.upper_bound(mp(p,4 * N)),g = k;
    --g;
    pii t2 = *k,t1 = *g;
    if(t1.se == p - 1 && t2.fi == p + 1) {
        S.erase(k);S.erase(g);
        S.insert(mp(t1.fi,t2.se));
    }
    else if(t1.se == p - 1) {
        S.erase(g);S.insert(mp(t1.fi,p));
    }
    else if(t2.fi == p + 1) {
        S.erase(k);S.insert(mp(p,t2.se));
    }
    else {
        S.insert(mp(p,p));
    }
    }
    B[p] = v;
}
int check(int p,pii t) {
    if(t.fi == 4 * N || t.fi == -4 * N) return 0;
    if((t.fi <= p && t.se >= p) || (t.fi <= p + N && t.se >= p + N)) return 0;
    else return min(t.fi - p,p + N - t.se);
}
int Query(int p) {
    if(S.size() == 2) return -1;
    auto k = S.lower_bound(mp(p + N / 2,4 * N)),g = k,h = k;
    --g;++h;
    int res = 0;
    pii t = *k;
    res = max(check(p,t),res);
    t = *g;
    res = max(check(p,t),res);
    if(h != S.end()) t = *h,res = max(check(p,t),res),++h;
    if(g != S.begin()) {
    --g;t = *g;res = max(check(p,t),res);
    }
    if(h != S.end()) {
    t = *h;res = max(check(p,t),res);
    }
    return res;
}
int calc(int a,int b,char op) {
    if(op == ‘*‘) return a * b % 10;
    else return (a + b) % 10;
}
void Solve() {
    read(N);read(M);
    S.insert(mp(4 * N,4 * N));
    S.insert(mp(-4 * N,-4 * N));
    for(int i = 1 ; i <= N ; ++i) {
    read(A[i]);scanf("%s",ch[i] + 1);
    }
    A[0] = A[N];
    for(int i = 1 ; i <= N ; ++i) {
    B[i] = calc(A[i],A[i - 1],ch[i][1]);
    }
    for(int i = 1 ; i <= N ; ++i) B[i + N] = B[i],A[i + N] = A[i];
    int p = -1;
    for(int i = 1 ; i <= 2 * N ; ++i) {
    if(B[i] == 0) {
        if(p == -1) p = i;
    }
    else if(p != -1) {
        S.insert(mp(p,i - 1));
        p = -1;
    }
    }
    if(p != -1) S.insert(mp(p,2 * N));
    int op,num;char s[5];
    for(int i = 1 ; i <= M ; ++i) {
    read(op);
    if(op == 1) {
        read(p);read(num);scanf("%s",s + 1);
        ++p;
        if(p == N) {
        Change(1,calc(A[1],num,ch[1][1]));
        Change(N + 1,calc(A[1],num,ch[1][1]));
        }
        else {
        Change(p + 1,calc(A[p + 1],num,ch[p + 1][1]));
        Change(p + 1 + N,calc(A[p + 1],num,ch[p + 1][1]));
        }
        A[p] = A[p + N] = num;ch[p][1] = s[1];
        Change(p,calc(A[p],A[p - 1],s[1]));
        Change(p + N,calc(A[p],A[p - 1],s[1]));
        if(p == N) A[0] = num;
    }
    else {
        read(p);++p;
        int t = B[p];
        Change(p,A[p]);
        out(Query(p));enter;
        Change(p,t);
    }
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

【LOJ】#2010. 「SCOI2015」小凸解密碼