• Bzoj4653: [Noi2016]区间


    题目

    传送门

    Sol

    把区间按长度升序排序,双端点+离散化+线段树区间最大值

    线段树标记永久化美滋滋

    # 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 _(5e5 + 1);
    
    IL ll Read(){
        RG ll 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, ql[_], qr[_], id[_], o[_ << 1], cnt, ans = 2e9, idl[_], idr[_], val[_];
    int mx[_ << 3], tag[_ << 3];
    
    IL bool Cmp(RG int x, RG int y){  return qr[x] - ql[x] < qr[y] - ql[y];  }
    
    IL void Modify(RG int x, RG int l, RG int r, RG int L, RG int R, RG int d){
    	if(L <= l && R >= r){  mx[x] += d; tag[x] += d; return;  }
    	RG int mid = (l + r) >> 1;
    	if(L <= mid) Modify(x << 1, l, mid, L, R, d);
    	if(R > mid) Modify(x << 1 | 1, mid + 1, r, L, R, d);
    	mx[x] = max(mx[x << 1], mx[x << 1 | 1]) + tag[x];
    }
    
    int main(RG int argc, RG char* argv[]){
    	n = Read(); m = Read();
    	for(RG int i = 1; i <= n; ++i){
    		ql[i] = Read(); qr[i] = Read(); id[i] = i;
    		o[++cnt] = ql[i]; o[++cnt] = qr[i];
    	}
    	sort(o + 1, o + cnt + 1); sort(id + 1, id + n + 1, Cmp);
    	cnt = unique(o + 1, o + cnt + 1) - o - 1;
    	for(RG int i = 1; i <= n; ++i){
    		val[i] = qr[id[i]] - ql[id[i]] + 1;
    		idl[i] = lower_bound(o + 1, o + cnt + 1, ql[id[i]]) - o;
    		idr[i] = lower_bound(o + 1, o + cnt + 1, qr[id[i]]) - o;
    	}
    	for(RG int i = 1, j = 0; i <= n; ++i){
    		while(j < n && mx[1] < m) ++j, Modify(1, 1, cnt, idl[j], idr[j], 1);
    		if(mx[1] < m) break;
    		ans = min(ans, val[j] - val[i]);
    		Modify(1, 1, cnt, idl[i], idr[i], -1);
    	}
    	printf("%d
    ", ans == 2e9 ? -1 : ans);
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8306162.html
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