1651: [Usaco2006 Feb]专用牛棚[经典] 多个区间不相交

1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 1031  Solved: 582
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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

4


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

第一种方法是直接用线段树暴力区间加1

 #include<bits/stdc++.h>using namespace std;typedef long long ll;const int INF = 1e9;  inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;} const int N = 1000005; struct {    int l,r,val,tag;}t[3000005];    void pushdown(int k){    //向下传递tag    if(t[k].l==t[k].r) return;     int tag = t[k].tag;    t[k].tag=0;     if(tag)    {        t[k<<1].tag =t[k<<1].tag+tag;        t[k<<1|1].tag=t[k<<1|1].tag+tag;        t[k<<1].val=t[k<<1].val+tag;        t[k<<1|1].val=t[k<<1|1].val+tag;    }} void build(int k,int l,int r){    t[k].l = l;    t[k].r = r;    if(l==r) return ;    int mid = (l+r)>>1;    build(k<<1,l,mid);    build(k<<1|1,mid+1,r);}  void update(int k,int x,int y,int val){    pushdown(k);     int l = t[k].l;    int r = t[k].r;     //更新最大值    if(l==x&&r==y)    {        t[k].tag ++;        t[k].val ++;        return ;    }     //分段更新    int mid = (l+r)>>1;    if(y<=mid)        update(k<<1,x,y,val);    else if(x>mid)        update(k<<1|1,x,y,val);    else    {        update(k<<1,x,mid,val);        update(k<<1|1,mid+1,y,val);    }     t[k].val=max(t[k<<1].val,t[k<<1|1].val);}  int query(int k,int x){    pushdown(k);    int l = t[k].l;    int r = t[k].r;    if(l==r)    {        return t[k].val;    }    int mid = (l+r)>>1;    if(x<=mid)    {        return query(k<<1,x);    }    else    {        return query(k<<1|1,x);    }}   int n;  int l,r; int main(){//    freopen("data.txt","r",stdin);    ios_base::sync_with_stdio(false);    cin >> n;    build(1,1,1000000);    for(int i=0;i<n;i++)    {        cin >> l>>r;        update(1,l,r,1);    }    pushdown(1);    cout << t[1].val<<endl;    return 0;}

第二种是用优先队列,优先右边值最小的

#include<bits/stdc++.h>using namespace std;typedef long long ll;const int INF = 1e9;inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}struct node {    int l;    int r;    bool operator < (const node &x) const    {        return r>x.r;    }};bool cmp(node x,node y){    return x.l<y.l;}node nt;node cur;int n;vector<node> a;int main(){//    freopen("data.txt","r",stdin);//    ios_base::sync_with_stdio(false);    n=read();    for(int i=0;i<n;i++)    {        nt.l=read();        nt.r=read();        a.push_back(nt);    }    sort(a.begin(),a.end(),cmp);    priority_queue<node> q;    q.push(a[0]);    for(int i=1;i<n;i++)    {        cur = q.top();        if(cur.r<a[i].l)        {            q.pop();        }        q.push(a[i]);    }    printf("%d\n",q.size());    return 0;}