POJ3190 Stall Reservations 【贪婪】

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3106		Accepted: 1117		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

Sample Output

Hint

Explanation of the sample:

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.

Source

USACO 2006 February Silver

/*
** 用堆维护的贪心题。先依照開始时间排序。再将牛依次放入堆里。放入之前更新堆顶元素。
** TLE到吐。。。
*/

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>

#define maxn 50010
using namespace std;

struct Node2 {
    int num, u, v;
    friend bool operator<(const Node2& a, const Node2& b) {
        return a.v > b.v;
    } 
} cow[maxn];
int ans[maxn];

bool cmp(const Node2& a, const Node2& b) {
    return a.u < b.u;
}

int main() {
    int N, i, j, sum, u, v, flag;
    Node2 tmp;
    while(scanf("%d", &N) == 1) {
        sum = 0;
        for(i = 0; i < N; ++i) {
            scanf("%d%d", &u, &v);
            cow[i].num = i + 1;
            cow[i].u = u;
            cow[i].v = v;
            ans[i + 1] = 0;
        }
        sort(cow, cow + N, cmp);

        priority_queue<Node2> PQ;
        PQ.push(cow[0]);
        ans[cow[0].num] = ++sum;

        for(i = 1; i < N; ++i) {
            tmp = PQ.top();
            if(cow[i].u > tmp.v) {
                tmp.v = cow[i].v;
                ans[cow[i].num] = ans[tmp.num];
                PQ.pop(); PQ.push(tmp);
            } else {
                ans[cow[i].num] = ++sum;
                PQ.push(cow[i]);
            }
        }

        printf("%d
", sum);
        for(i = 1; i <= N; ++i)
            printf("%d
", ans[i]);
    }
    return 0;
}

超时代码1：

#include <stdio.h>
#include <string.h>
#include <algorithm>

#define maxn 50010
using std::sort;

struct Node {
    int u, v;    
} E[maxn];
struct Node2 {
    int num, u, v;
} cow[maxn];
int ans[maxn];

bool cmp(const Node2& a, const Node2& b) {
    return a.u < b.u;
}

int main() {
    int N, i, j, sum, u, v;
    while(scanf("%d", &N) == 1) {
        sum = 0;
        for(i = 0; i < N; ++i) {
            scanf("%d%d", &u, &v);
            cow[i].num = i + 1;
            cow[i].u = u;
            cow[i].v = v;
        }
        sort(cow, cow + N, cmp);

        for(i = 0; i < N; ++i) {
            E[i].v = 0;
            for(j = 0; j <= i; ++j) {
                if(cow[i].u > E[j].v) {
                    if(!E[j].v) ++sum;
                    E[j].v = cow[i].v;
                    E[j].u = cow[i].u;
                    ans[cow[i].num] = j + 1;
                    break;
                }
            }
        }

        printf("%d
", sum);
        for(i = 1; i <= N; ++i)
            printf("%d
", ans[i]);
    }
    return 0;
}

超时代码2：

#include <stdio.h>
#include <string.h>
#include <algorithm>

#define maxn 50010
using std::sort;

struct Node {
    int u, v;    
} E[maxn];
struct Node2 {
    int num, u, v;
} cow[maxn];
int ans[maxn];

bool cmp(const Node2& a, const Node2& b) {
    return a.u < b.u;
}

int main() {
    int N, i, j, sum, u, v, flag;
    while(scanf("%d", &N) == 1) {
        sum = 0;
        for(i = 0; i < N; ++i) {
            scanf("%d%d", &u, &v);
            cow[i].num = i + 1;
            cow[i].u = u;
            cow[i].v = v;
            ans[i + 1] = 0;
        }
        sort(cow, cow + N, cmp);

        for(i = 0; i < N; ++i) {
            if(ans[cow[i].num]) continue;
            ans[cow[i].num] = ++sum;
            flag = cow[i].v;
            for(j = i + 1; j < N; ++j)
                if(!ans[cow[j].num] && cow[j].u > flag) {
                    flag = cow[j].v;
                    ans[cow[j].num] = sum;
                }
        }

        printf("%d
", sum);
        for(i = 1; i <= N; ++i)
            printf("%d
", ans[i]);
    }
    return 0;
}