【BZOJ】1651: [Usaco2006 Feb]Stall Reservations 专用牛棚（线段树/前缀和 + 差分）

http://www.lydsy.com/JudgeOnline/problem.php?id=1651

很奇妙。。

我们发现，每一时刻的重叠数选最大的就是答案。。。。

orz

那么我们可以线段树维护每个点的次数。。。

然后就ok了。。

第二种做法：用前缀和来维护即可。。。

线段树：

#include <cstdio>

#include <cstring>

#include <cmath>

#include <string>

#include <iostream>

#include <algorithm>

#include <queue>

using namespace std;

#define rep(i, n) for(int i=0; i<(n); ++i)

#define for1(i,a,n) for(int i=(a);i<=(n);++i)

#define for2(i,a,n) for(int i=(a);i<(n);++i)

#define for3(i,a,n) for(int i=(a);i>=(n);--i)

#define for4(i,a,n) for(int i=(a);i>(n);--i)

#define CC(i,a) memset(i,a,sizeof(i))

#define read(a) a=getint()

#define print(a) printf("%d", a)

#define dbg(x) cout << #x << " = " << x << endl

#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }

inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

inline const int max(const int &a, const int &b) { return a>b?a:b; }

inline const int min(const int &a, const int &b) { return a<b?a:b; }

#define lc x<<1

#define rc x<<1|1

#define lson l, m, lc

#define rson m+1, r, rc

#define MID (l+r)>>1

struct nod { int mx, tag; }t[4000005];

int n;

void pushdown(int x) {

	int g=t[x].tag; t[x].tag=0;

	t[lc].mx+=g; t[lc].tag+=g;

	t[rc].mx+=g; t[rc].tag+=g;

}

void pushup(int x) { t[x].mx=max(t[lc].mx, t[rc].mx); }

void update(int l, int r, int x, int L, int R) {

	if(L<=l && r<=R) { ++t[x].mx; ++t[x].tag; return; }

	pushdown(x);

	int m=MID;

	if(L<=m) update(lson, L, R); if(m<R) update(rson, L, R);

	pushup(x);

}

int main() {

	read(n);

	for1(i, 1, n) {

		int l=getint(), r=getint();

		update(1, 1000000, 1, l, r);

	}

	print(t[1].mx);

	return 0;

}

前缀和：

#include <cstdio>

#include <cstring>

#include <cmath>

#include <string>

#include <iostream>

#include <algorithm>

#include <queue>

using namespace std;

#define rep(i, n) for(int i=0; i<(n); ++i)

#define for1(i,a,n) for(int i=(a);i<=(n);++i)

#define for2(i,a,n) for(int i=(a);i<(n);++i)

#define for3(i,a,n) for(int i=(a);i>=(n);--i)

#define for4(i,a,n) for(int i=(a);i>(n);--i)

#define CC(i,a) memset(i,a,sizeof(i))

#define read(a) a=getint()

#define print(a) printf("%d", a)

#define dbg(x) cout << #x << " = " << x << endl

#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }

inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }

inline const int max(const int &a, const int &b) { return a>b?a:b; }

inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int M=1000005;

int n, t[M], sum, mx, ans;

int main() {

	read(n);

	for1(i, 1, n) {

		int x=getint(), y=getint();

		++t[x]; --t[y+1]; mx=max(mx, y);

	}

	for1(i, 1, mx) {

		sum+=t[i];

		if(sum>ans) ans=sum;

	}

	print(ans);

	return 0;

}

Description

Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. Of course, no cow will share such a private moment with other cows. Help FJ by determining: * The minimum number of stalls required in the barn so that each cow can have her private milking period * An assignment of cows to these stalls over time

有N头牛,每头牛有个喝水时间,这段时间它将专用一个Stall 现在给出每头牛的喝水时间段,问至少要多少个Stall才能满足它们的要求

Input

* Line 1: A single integer, N

* Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

Output

* Line 1: The minimum number of stalls the barn must have.

* Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.

Sample Input

5
1 10
2 4
3 6
5 8
4 7

Sample Output

OUTPUT DETAILS:

Here's a graphical schedule for this output:

Time 1 2 3 4 5 6 7 8 9 10
Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>
Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..
Stall 3 .. .. c3>>>>>>>>> .. .. .. ..
Stall 4 .. .. .. c5>>>>>>>>> .. .. ..

Other outputs using the same number of stalls are possible.

HINT

不妨试下这个数据,对于按结束点SORT,再GREEDY的做法 1 3 5 7 6 9 10 11 8 12 4 13 正确的输出应该是3

Source

Silver

巴特西