POJ 2376 Cleaning Shifts（贪心）

Description

Input

Output

Sample Input

3 10
1 7
3 6
6 10

Sample Output

2

最小代价的区间覆盖。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<map>
using namespace std;
typedef long long LL;
const int maxn=25000+100;
struct node{
    int st,ed;
}e[maxn];
int n,m;
int cmp(node l1,node l2)
{
    if(l1.st==l2.st)
        return  l1.ed>l2.ed;
    return l1.st<l2.st;
}
int main()
{
    int ed,temp;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<n;i++)
            scanf("%d%d",&e[i].st,&e[i].ed);
        sort(e,e+n,cmp);
        if(e[0].st!=1)
        {
            printf("-1\n");
            continue;
        }
        int ans=1;
        ed=e[0].ed;
        temp=0;
        for(int i=1;i<n;i++)
        {
            if(e[i].st>ed+1)
            {
                ed=temp;
                ans++;
            }
            if(e[i].st<=ed+1)
            {
                if(e[i].ed>temp)
                    temp=e[i].ed;
                if(e[i].ed==m)
                {
                    ans++;
                    ed=m;
                    break;
                }
            }
        }
        if(ed==m)   printf("%d\n",ans);
        else   printf("-1\n");
    }
    return 0;
}