POJ 3190 Stall Reservations（贪心）

Stall Reservations

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3590 Accepted: 1284 Special Judge

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

Many answers are correct for each test dataset; a program will grade your answer.

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

51 102 43 65 84 7

Sample Output

412324

Hint

Explanation of the sample: 

Here's a graphical schedule for this output: 

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

Other outputs using the same number of stalls are possible.

Source

USACO 2006 February Silver

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define epst 1e-8typedef __int64 ll;#define fre(i,a,b)  for(i = a; i <b; i++)#define free(i,b,a) for(i = b; i >= a;i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define ssf(n)      scanf("%s", n)#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define bug         pf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 50005struct stud{  int x,y;  int pos;  bool operator <(const stud d) const  {return y>d.y;  }}f[N];int ans[N];int k;priority_queue<stud>q;int n;int cmp(stud a,stud b){if(a.x==b.x) return a.y<b.y;return a.x<b.x;}void solve(){int i,j;stud cur;while(!q.empty())q.pop();for(i=1;i<=n;i++){if(q.empty()){ans[f[i].pos]=++k;q.push(f[i]);continue;}        cur=q.top();        if(cur.y<f[i].x){ans[f[i].pos]=ans[cur.pos];q.pop();}        else  ans[f[i].pos]=++k;        q.push(f[i]);}}int main(){int i,j;while(~sf(n)){for(i=1;i<=n;i++){sff(f[i].x,f[i].y);    f[i].pos=i;}sort(f+1,f+n+1,cmp);k=0;solve();pf("%d\n",k);for(i=1;i<=n;i++)pf("%d\n",ans[i]);}    return 0;}