[贪心]uva10020 Minimal coverage

Minimal coverage

The Problem

Given several segments of line (int the X axis) with coordinates [L_i,R_i]. You are to choose the minimal amount of them, such they would completely cover the segment [0,M].

The Input

The first line is the number of test cases, followed by a blank line.

Each test case in the input should contains an integer M(1<=M<=5000), followed by pairs "L_i R_i"(|L_i|, |R_i|<=50000, i<=100000), each on a separate line. Each test case of input is terminated by pair "0 0".

Each test case will be separated by a single line.

The Output

For each test case, in the first line of output your programm should print the minimal number of line segments which can cover segment [0,M]. In the following lines, the coordinates of segments, sorted by their left end (L_i), should be printed in the same format as in the input. Pair "0 0" should not be printed. If [0,M] can not be covered by given line segments, your programm should print "0"(without quotes).

Print a blank line between the outputs for two consecutive test cases.

Sample Input

2

1
-1 0
-5 -3
2 5
0 0

1
-1 0
0 1
0 0

Sample Output

0

1
0 1

Alex Gevak
September 10, 2000 (Revised 2-10-00, Antonio Sanchez)

题意：给出一个数字M,再给出一些区间，求出使用最少的区间使这些区间覆盖0-M。

思路：典型的贪心题目，首先预处理，把和所给的范围不相关的区间过滤掉。再在这些区间中选择使右边的区间能到达最远的范围。

#include<iostream>
#include<algorithm>
#include<vector>

using namespace std;

class Node
{
public:
    int x,y;
}node[100005];

vector<Node> st;

bool cmp(Node r,Node t)
{
    return r.x<t.x;
}

int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        int m,num=0;
        st.clear();
        cin>>m;
        int left,right;
        while(cin>>left>>right&&(left!=0||right!=0))
        {
            if(left<m&&right>0)
            {
                node[num].x=left;
                node[num++].y=right;
            }
        }
        sort(node,node+num,cmp);
        int start=0,tag=1,tag1;
        int i,j,k,maxlen;
        while(start<m)
        {
            maxlen=start,tag1=0;
            for(i=0;i<num;i++)
            {
                if(node[i].y>maxlen&&node[i].x<=start) maxlen=node[i].y,k=i,tag1=1;
                else if(node[i].x>start) break;
            }
            if(tag1==0)
            {
                tag=0;
                break;
            }
            start=node[k].y;
            st.push_back(node[k]);
        }
        if(tag==0) cout<<"0"<<endl;
        else
        {
            cout<<st.size()<<endl;
            for(i=0;i<st.size();i++)
                cout<<st[i].x<<" "<<st[i].y<<endl;
        }
        if(n!=1) cout<<endl;
    }
    return 0;
}