【BZOJ3809】Gty的二逼妹子序列【莫队算法】【分块】

【题目链接】

一开始写了个莫队+树状数组，T飞了。

需要对权值分块，然后分块查询。

/* Telekinetic Forest Guard */
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int maxn = 100005, maxm = 1000005;

int n, m, num[maxn], belong[maxn], pre[maxn], last[maxn], size, ans[maxm], vis[maxn], blo[maxn];

struct _que {
int id, l, r, a, b;

bool operator < (const _que &x) const {
return belong[l] != belong[x.l] ? belong[l] < belong[x.l] : r < x.r;
}
} que[maxm];

inline int iread() {
int f = 1, x = 0; char ch = getchar();
for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
return f * x;
}

inline int query(int a, int b) {
int res = 0, L = belong[a], R = belong[b];
for(int i = L + 1; i < R; i++) res += blo[i];
if(L == R) {
for(int i = a; i <= b; i++) if(vis[i]) res++;
} else {
for(int i = a; i <= last[a]; i++) if(vis[i]) res++;
for(int i = pre[b]; i <= b; i++) if(vis[i]) res++;
}
return res;
}

inline void add(int x) {
x = num[x];
if(!vis[x]) blo[belong[x]]++;
vis[x]++;
}

inline void del(int x) {
x = num[x];
vis[x]--;
if(!vis[x]) blo[belong[x]]--;
}

int main() {
n = iread(); m = iread(); size = (int)sqrt(n / 2);
for(int i = 1; i <= n; i++) num[i] = iread(), belong[i] = (i - 1) / size + 1, pre[i] = (belong[i] - 1) * size + 1, last[i] = belong[i] * size;
for(int i = 1; i <= m; i++) que[i] = (_que){i, iread(), iread(), iread(), iread()};

sort(que + 1, que + 1 + m);

int l = 1, r = 0;
for(int i = 1; i <= m; i++) {
for(; que[i].l < l; add(--l));
for(; r < que[i].r; add(++r));
for(; l < que[i].l; del(l++));
for(; que[i].r < r; del(r--));
ans[que[i].id] = query(que[i].a, que[i].b);
}

for(int i = 1; i <= m; i++) printf("%d\n", ans[i]);
return 0;
}