C- interval coverage (does not support C ++ 11)

C- interval coverage (does not support C ++ 11)

One, Title Description

There are a number line n (1 <= n <= 25000) a closed interval [ai, bi], selected to cover a range of as little as specified segment [1, t] (1 < = t <= 1,000,000).
Covering the entire point, i.e., (1,2) + (3,4) may cover (1,4).
The impossible Output -1

Entry

   第一行：N和T
   第二行至N+1行: 每一行一个闭区间。

Export

选择的区间的数目，不可能办到输出-1

Sample input

3 10
1 7
3 6
6 10

Sample Output

2

prompt

   这道题输入数据很多，请用scanf而不是cin

Second, ideas and algorithms

The minimum number of sections required to cover a certain period interval, this algorithm is the main problem, first of all sorted interval (interval to a first point of the left ascending keywords to the right end of the second section ascending keyword), then [1 , t] both sides of the cut section.
Thereafter, as the core: each of a selected interval, the cover portion has been cut off are, in the cover section can continue down, the selected maximum interval until [1, t] are covered completed.
Must pay attention to are: an integer coverage requirements, so there are requirements +1.
2. Select a starting point of the first interval must be 1, otherwise the sentence is impossible.
3. The key may exceed t, e.g. [1,12] Cover [1,10]

Third, code implementation

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;

struct period{
	int a,b;
	period(){}
	period(int aa,int bb){a=aa;		b=bb;}
	bool operator <(const period &p)const{    //重载<
		if(a!=p.a){return a<p.a;}
		else{return b<p.b;} 
	}
};

int main(){
	int N,T;
	vector<period> all(0);
	int ta,tb;
	//读入所有区间
	scanf("%d %d",&N,&T);
	for(int i=0;i<N;i++){
		scanf("%d %d",&ta,&tb);
		all.push_back(period(ta,tb));
	}
	sort(all.begin(),all.end());
	
	int s=all.size();
	//切掉[1,T]以外的部分 
	while(all.empty()==0&&all[s-1].a>T){
		all.pop_back();
		s--;
	}
	while(all.empty()==0&&all[0].a<1){
		vector<period>::iterator it=all.begin();
		all.erase(it);
		s--;
	}
	//开始选择区间
	if(all.empty()==1){printf("-1\n");	return 0;}
	int result=0;          //需要的区间总数
	if(all[0].a>1){printf("-1\n");	return 0;}
	int tmp=0,j=0;
	int latest=1,flag=0;
	while(latest<T)
	{
		int maxlen=0;  //目前长度最长区间长度 
		if(flag==0){  //flag只在第一次为0，处理第一个区间 
			flag=1;
			for(int i=0;(all[i].a==1)&&i<all.size();i++){
				tmp=all[i].b-1;
				if(maxlen<tmp){maxlen=tmp;	j=i;}
			}
			if(maxlen==0) {printf("-1\n");	return 0;}
			result++;	latest=all[j].b;
		}
		else{
			for(int i=0;(i<all.size())&&(all[i].a<=(latest+1));i++)//找当前左边界最长 
		    {	if(all[i].b>latest){
		    		tmp=all[i].b-latest;
					if(tmp>maxlen)
					{
						maxlen=tmp;
						j=i;
					}
				}				
			}
			if(maxlen==0) {printf("-1\n");	return 0;}//不存在覆盖 
			else{
				result++;
				latest=all[j].b;
			}
		} 
	}
	printf("%d\n",result);		
	return 0;
}

Fourth, experience and summary

1. Remember Remember! ! Heavy load <Be sure to include all cases! What returns are not equal when, what time and equal return, or will WA!
2. Time uncertain cycle boundary and other data, you can use test data, tracking data operation to determine the boundaries set correctly, how mistaken the change.
3. scanf time performance is better than cin, more recommended scanf, printf.
4. Greedy index is the most important part of the greedy algorithm, the correctness of the greedy algorithm dependent on the correctness of indicators, so greedy indicators must be carefully chosen!