• Bzoj4552: [Tjoi2016&Heoi2016]排序


    题目

    传送门

    Sol

    二分+线段树
    巧妙啊我怎么就没想到
    二分答案,把数分类,大于等于(mid)的为(1),小于的为(0)
    相当于给(01)序列排序,最后判断询问位置上是不是(1)
    线段树+lazy覆盖

    # include <bits/stdc++.h>
    # define RG register
    # define IL inline
    # define Fill(a, b) memset(a, b, sizeof(a))
    using namespace std;
    typedef long long ll;
    const int _(1e5 + 5);
    
    IL int Input(){
        RG int x = 0, z = 1; RG char c = getchar();
        for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
        for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
        return x * z;
    }
    
    int n, m, q, a[_], ql[_], qr[_], qo[_];
    int sum[_ << 2], cov[_ << 2];
    
    IL void Build(RG int x, RG int l, RG int r, RG int v){
    	sum[x] = 0, cov[x] = -1;
    	if(l == r){
    		sum[x] = a[l] >= v;
    		return;
    	}
    	RG int mid = (l + r) >> 1, ls = x << 1, rs = x << 1 | 1;
    	Build(ls, l, mid, v), Build(rs, mid + 1, r, v);
    	sum[x] = sum[ls] + sum[rs];
    }
    
    IL void Pushdown(RG int x, RG int l, RG int mid, RG int r){
    	RG int ls = x << 1, rs = x << 1 | 1;
    	cov[ls] = cov[rs] = cov[x];
    	sum[ls] = (mid - l + 1) * cov[x];
    	sum[rs] = (r - mid) * cov[x];
    	cov[x] = -1;
    }
    
    IL void Modify(RG int x, RG int l, RG int r, RG int L, RG int R, RG int v){
    	if(L <= l && R >= r){
    		sum[x] = (r - l + 1) * v;
    		cov[x] = v;
    		return;
    	}
    	RG int mid = (l + r) >> 1, ls = x << 1, rs = x << 1 | 1;
    	if(cov[x] != -1) Pushdown(x, l, mid, r);
    	if(L <= mid) Modify(ls, l, mid, L, R, v);
    	if(R > mid) Modify(rs, mid + 1, r, L, R, v);
    	sum[x] = sum[ls] + sum[rs];
    }
    
    IL int Query(RG int x, RG int l, RG int r, RG int L, RG int R){
    	if(cov[x] != -1) return cov[x] * (R - L + 1);
    	if(L <= l && R >= r) return sum[x];
    	RG int mid = (l + r) >> 1;
    	if(R <= mid) return Query(x << 1, l, mid, L, R);
    	if(L > mid) return Query(x << 1 | 1, mid + 1, r, L, R);
    	return Query(x << 1, l, mid, L, mid) + Query(x << 1 | 1, mid + 1, r, mid + 1, R);
    }
    
    IL bool Check(RG int mid){
    	Build(1, 1, n, mid);
    	for(RG int i = 1; i <= m; ++i){
    		RG int cnt = Query(1, 1, n, ql[i], qr[i]);
    		if(!cnt || cnt == qr[i] - ql[i] + 1) continue;
    		if(qo[i]){
    			Modify(1, 1, n, ql[i], ql[i] + cnt - 1, 1);
    			Modify(1, 1, n, ql[i] + cnt, qr[i], 0);
    		}
    		else{
    			Modify(1, 1, n, ql[i], qr[i] - cnt, 0);
    			Modify(1, 1, n, qr[i] - cnt + 1, qr[i], 1);
    		}
    	}
    	return Query(1, 1, n, q, q);
    }
    
    int main(RG int argc, RG char* argv[]){
    	n = Input(), m = Input();
    	for(RG int i = 1; i <= n; ++i) a[i] = Input();
    	for(RG int i = 1; i <= m; ++i) qo[i] = Input(), ql[i] = Input(), qr[i] = Input();
    	q = Input();
    	RG int l = 1, r = n, ans = 0;
    	while(l <= r){
    		RG int mid = (l + r) >> 1;
    		if(Check(mid)) ans = mid, l = mid + 1;
    		else r = mid - 1;
    	}
    	printf("%d
    ", ans);
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8469487.html
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