POJ2376贪心——求助--更新

POJ 2376 贪心

求助!!!
请问有缘看到此贴的人能给出帮助吗?帮看看我的代码为什么TLE了
大恩不言谢:
题目:https://vjudge.net/problem/POJ-2376

Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T.

Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval.

Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.
Input

  • Line 1: Two space-separated integers: N and T

  • Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.
    Output

  • Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.
    Sample Input
    3 10
    1 7
    3 6
    6 10
    Sample Output
    2
    Hint
    This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

INPUT DETAILS:

There are 3 cows and 10 shifts. Cow #1 can work shifts 1..7, cow #2 can work shifts 3..6, and cow #3 can work shifts 6..10.

OUTPUT DETAILS:

By selecting cows #1 and #3, all shifts are covered. There is no way to cover all the shifts using fewer than 2 cows.

题目大概意思就是给奶牛排班,n头奶牛把t时间排满,行就输出最小奶牛数,否则输出-1,相信做过的不在少数吧
一开始我代码少了个+1,然后WA,加上之后又TLE了,想问问怎么改,评论区应该可以用吧

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#include <iostream>
#include <cstdio>
using namespace std;
int main(){
int n,t;
int st[25005],end[25005];
scanf("%d%d",&n,&t);
for(int i=0;i<n;i++){
scanf("%d%d",&st[i],&end[i]);
}
int ans=0;
int fir,zd=0;
for(int i=0;i<n;i++){
if(st[i]==1){
fir=i;
zd=end[i];
break;
}
else if(i==n-1){
printf("-1\n");
return 0;
}
}
for(int i=0;i<n;i++){
if(end[i]>=t){
break;
}
else if(i==n-1){
printf("-1\n");
return 0;
}
}
for(int i=fir+1;i<n;i++){
if(st[i]==1){
if(end[i]>end[fir]){
fir=i;
zd=end[fir];
}
}
}
ans++;
while(zd<t){
int ms=-100000;
int pos;
for(int i=0;i<n;i++){
if(st[i]<=zd+1&&end[i]>zd){
if(end[i]-zd>ms){
ms=end[i]-zd;
pos=i;
}
}
}
if(ms==-100000){
printf("-1\n");
return 0;
}
zd=end[pos];
ans++;
}
printf("%d\n",ans);
return 0;
}

以上为暑期的错误代码
今天-20180928
又重新审视了一次(看了下题解
有点费劲,但是最后还是理解了

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/*Sample Input
3 10
1 7
3 6
6 10
Sample Output
2*/

#include <iostream>
#include <stack>
#include <algorithm>
using namespace std;
const int maxn=25005;
struct cow
{
int l,r;
}c[maxn];
bool cmp(cow a,cow b){
if(a.l==b.l)return a.r<b.r;
return a.l<b.l;
}
int num[1000005];
int main(){
ios::sync_with_stdio(false);
int n,t;
cin>>n>>t;
for(int i=0;i<n;i++){
cin>>c[i].l>>c[i].r;
}
sort(c,c+n,cmp);
int num=0;
int ans=0;
int end=0;
while(end<t){
int fir=end+1;//待覆盖的起点
for(int i=ans;i<n;i++){
if(c[i].l<=fir&&c[i].r>=fir){
end=max(c[i].r,end);
}
else if(c[i].l>fir){
ans=i;
break;
}
}
if(fir>end)break;
else num++;
}
if(end==t)cout<<num<<endl;
else cout<<"-1"<<endl;
return 0;
}

其实主要就是这段代码

就是更新end的值,一直到t,期间没选中一次奶牛,num++;然后输出 ok

感谢
coswindy

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