区间调度问题

Description

有 n 项工作,每项工作分别在 si 时间开始,在 ti 时间结束。对于每项工作,你都可以选择参与否。如果选择了参与,那么自始至终都必须全程参与。此外,参与工作的时间段不能重叠(既是是开始的瞬间和结束的瞬间重叠也是不允许的)

你的目标是参与尽可能多的工作,那么最多能参与多少项工作呢?

Input

第一行输入一个数 n (1 ≤ n ≤ 100000),表示工作的数量
接下来的 n 行,第 1 + i (1 ≤ i ≤ n) 行包含两个数 si, ti (1 ≤ si ≤ ti ≤ 10^9),分别为起始时间和结束时间。

Output

输出为一行,包含一个整数,为能参与的工作数的最大值

Sample Input

5
1 3
2 5
4 7
6 9
8 10

Sample Output

3

Hint

每一次都选取结束时间最早的事件。因为与其他选择方案相比,该算法的选择方案在选取了相同数量的更早开始的工作时,
其最终结束时间不会比其他方案的更晚。所以,不存在选取更多工作的选择方案。
通过定义 compare 函数实现 sort() 的自定义排序规则。

题解

#include 
#include 
#include 
#include 
#include 
#include 
#define INF 100100
using namespace std;
int sum = 0;

struct node
{
    int start, end;
}arr[INF];

bool cmp(const node& i, const node& j)
{
    if (i.end > j.end)
        return 0;
    else
        if (i.end == j.end && i.start > j.start)
            return 0;
    return 1;
}

inline int find(int addr, int n)
{
    for (int i = addr + 1; i < n; i++)
        if (arr[i].start > arr[addr].end){
            return i;
        }
    return n+1;
}

int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d%d", &arr[i].start, &arr[i].end);
    sort(arr,arr+n,cmp);
    for (int addr = 0; addr < n; addr = find(addr, n))
        sum++;
    printf("%d\n", sum);
    return 0;
}

[POJ 3617]Best Cow Line

Description

FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual”Farmer of the Year” competition. In this contest every farmer arranges his cows in a line and herds them past the judges.

The contest organizers adopted a new registration scheme this year: simply register the initial letter of every cow in the order they will appear (i.e., If FJ takes Bessie, Sylvia, and Dora in that order he just registers BSD). After the registration phase ends, every group is judged in increasing lexicographic order according to the string of the initials of the cows’ names.

FJ is very busy this year and has to hurry back to his farm, so he wants to be judged as early as possible. He decides to rearrange his cows, who have already lined up, before registering them.

FJ marks a location for a new line of the competing cows. He then proceeds to marshal the cows from the old line to the new one by repeatedly sending either the first or last cow in the (remainder of the) original line to the end of the new line. When he’s finished, FJ takes his cows for registration in this new order.

Given the initial order of his cows, determine the least lexicographic string of initials he can make this way.

Input

  • Line 1: A single integer: N
  • Lines 2~N+1: Line i+1 contains a single initial (‘A’..’Z’) of the cow in the ith position in the original line

Output

The least lexicographic string he can make. Every line (except perhaps the last one) contains the initials of 80 cows (‘A’..’Z’) in the new line.

Sample Input

6
A
C
D
B
C
B

Sample Output

ABCBCD

Hint

使用贪心算法,不断将较小的字符放入新字符串的末尾,这样我们就能构造一个字典序最小的串。对于首尾字符相同的情景,取哪个都可以。
注意题目要求,每 80 字符换行。

题解

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <cstring>
//#include <regex>
#define online
#define INF 2010
using namespace std;

int n, head, tail;
char str[INF];
string str2;
int chk_head,chk_tail;
inline bool check()
{
    #ifdef offline
    printf("new check\n");
    #endif
    chk_head = head;
    chk_tail = tail;
    while (chk_head <= chk_tail)
    {
        if (str[chk_head] == str[chk_tail])
        {
            #ifdef offline
            printf("str[chk_head] %d %c== str[chk_tail] %d %c\n", chk_head, str[chk_head], chk_tail, str[chk_tail]);
            #endif
            chk_head++;
            chk_tail--;
        }
        else
        {
            if (str[chk_head] > str[chk_tail])
            {
                str2 = str2 + str[tail];
                tail--;
                #ifdef offline
                printf("check end at tail\n");
                #endif
                return 1;//tail
            }
            else
            {
                str2 = str2 + str[head];
                head++;
                #ifdef offline
                printf("check end at head\n");
                #endif
                return 0;//head
            }
        }
    }
    str2 = str2 + str[tail];
    tail--;
    #ifdef offline
    printf("check end at outside\n");
    #endif
}
inline bool solve()
{
    #ifdef offline
    printf("new solve\n");
    #endif
    if (str[head] > str[tail]){
        str2 = str2 + str[tail];
        tail--;
        #ifdef offline
        printf("solve 1\n");
        #endif
    }
    else
    {
        #ifdef offline
        printf("solve 0\n");
        #endif

        if (str[head] == str[tail] && head != tail){
            #ifdef offline
            printf("head == tail ==  %d\n", head);
            #endif
            check();
        }
        else
        {
            str2 = str2 + str[head];
            head++;
        }
    }
    if (head > tail)
        return 0;
    else
        solve();
}

int main()
{
    //srand((unsigned int)(time(NULL)));
    //freopen("","r",stdin);
    //freopen("","W",stdout);
    scanf("%d", &n);
    head = 0;
    tail = n-1;
    for (int i = 0; i < n; i++)
        scanf("%c%c", &str[i], &str[i]);
    solve();
    //cout<<str2;//poisonous output format
    for (int i = 0; i < str2.length(); i++)
    {
        if ((i + 1) % 80 == 0){
            if (i + 1 == n)
                printf("%c", str2[i]);
            else
                printf("%c\n", str2[i]);
        }
        else
            printf("%c", str2[i]);
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}