UVA 10020 - Minimal coverage 解题心得

原题：

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

分析：经典的贪心题，记得贪心一点就能做对了~

代码：

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

struct interval
{
    int x, y;

}itvl[100006], max_end[100006];

bool cmp(const interval &a, const interval &b)
{
    return a.x < b.x;
}

int main()
{
    int kase;
    cin >> kase;
    getchar();
    while (kase--){
        int ok = 0;
        int m,n=0;
        int cnt = 0;
        cin >> m;
        while (scanf("%d%d", &itvl[n].x, &itvl[n].y) && (itvl[n].x != 0 || itvl[n].y != 0)){
            ++n;
        }
        sort(itvl, itvl + n,cmp);

        int beg = 0, end = m;
        while (beg < end){
            max_end[cnt].y = 0;
            for (int i = 0; i < n; i++){
                if (itvl[i].x <= beg){
                    if (itvl[i].y>max_end[cnt].y)    {
                        max_end[cnt] = itvl[i];
                        ok = 1;
                    }
                }
                else{
                    break;
                }
            }
            if (ok == 0)        break;
            beg = max_end[cnt].y;
            cnt++;
        }

        //                        输出

        if (ok == 1){
            cout << cnt << endl;
            for (int i = 0; i < cnt; i++){
                cout << max_end[i].x << ‘ ‘ << max_end[i].y << endl;
            }
        }
        else{
            cout <<‘0‘<< endl;
        }
        if (kase != 0)            cout << endl;

    }
    return 0;
}