BZOJ4653 尺取法 + 线段树

https://www.lydsy.com/JudgeOnline/problem.php?id=4653

 

首先很容易想到离散之后排序，用线段树或者树状数组去维护。

问题在于按照什么排序，如果按照左端点右端点排序，线段树就需要维护最大值最小值和区间和等等信息，在区间和超过M之后最大值就变为了K大到最大的信息，不但麻烦而且难以下手。

所以想到直接按照区间长度排序，仅仅维护一个区间最值，代表这个区间里最大的点被覆盖了多少次，然后用一种尺取的思想，使得线段树里的最大值一值不超过M，如果超过了就在尾部开始删除直到M以下，与此同时更新最大值。

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 5e5 + 10;
const int maxm = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
int Hash[maxn * 2];
int len[maxn];
PII line[maxn];
struct Tree{
    int l,r,lazy,Max;
}tree[maxn << 3];
void Build(int t,int l,int r){
    tree[t].l = l; tree[t].r = r;
    tree[t].lazy = tree[t].Max = 0;
    if(l == r) return;
    int m = (l + r) >> 1;
    Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r);
}
void Pushdown(int t){
    if(tree[t].lazy){
        tree[t << 1].lazy += tree[t].lazy;
        tree[t << 1 | 1].lazy += tree[t].lazy;
        tree[t << 1].Max += tree[t].lazy;
        tree[t << 1 | 1].Max += tree[t].lazy;
        tree[t].lazy = 0;
    }
}
void Pushup(int t){
    tree[t].Max = max(tree[t << 1].Max,tree[t << 1 | 1].Max);
}
void update(int t,int l,int r,int w){
    if(l <= tree[t].l && tree[t].r <= r){
        tree[t].Max += w;
        tree[t].lazy += w;
        return;
    }
    Pushdown(t);
    int m = (tree[t].l + tree[t].r) >> 1;
    if(r <= m) update(t << 1,l,r,w);
    else if(l > m) update(t << 1 | 1,l,r,w);
    else{
        update(t << 1,l,m,w); update(t << 1 | 1,m + 1,r,w);
    }
    Pushup(t);
}
bool cmp(PII a,PII b){
    return a.se - a.fi < b.se - b.fi;
}
int main(){
    Sca2(N,M);
    int cnt = 0;
    for(int i = 1; i <= N; i ++){
        line[i].fi = read(); line[i].se = read();
        Hash[++cnt] = line[i].fi; Hash[++cnt] = line[i].se;
    }
    sort(Hash + 1,Hash + 1 + cnt);
    sort(line + 1,line + 1 + N,cmp);
    cnt = unique(Hash + 1,Hash + 1 + cnt) - Hash - 1;
    for(int i = 1; i <= N ; i ++){
        len[i] = line[i].se - line[i].fi;
        line[i].fi = lower_bound(Hash + 1,Hash + 1 + cnt,line[i].fi) - Hash;
        line[i].se = lower_bound(Hash + 1,Hash + 1 + cnt,line[i].se) - Hash;
    }
    Build(1,1,cnt);
    int tail = 1;
    int ans = INF;
    for(int i = 1; i <= N ; i ++){
        update(1,line[i].fi,line[i].se,1);
        while(tree[1].Max >= M){
            ans = min(ans,len[i] - len[tail]);
            update(1,line[tail].fi,line[tail].se,-1);
            tail++;
        }
    }
    if(ans == INF) ans = -1;
    Pri(ans);
    return 0;
}