【区间覆盖问题】uva 10020

可以说是区间覆盖问题的例题...

Note:

区间包含+排序扫描；

　　要求覆盖区间[s, t];

　　1、把各区间按照Left从小到大排序，如果区间1的起点大于s，则无解（因为其他区间的左起点更大）；否则选择起点在s的最长区间;

　　2、选择区间[li, ri]后，新的起点应更新为ri，并且忽略所有区间在ri之前的部分；

 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


代码如下

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100010;
struct Seg
{
    int L, R;
    bool operator < (const Seg &a) const
    {
        if(L != a.L) return L < a.L;
        else return R > a.R;
    }
} seg[maxn], ans[maxn];
void solve(int n, int m)
{
    int cnt = 0, l = 0, r = 0;//起点l，起点l最长区间的右端点r
    if(seg[0].L > l)
    {
        printf("0
");
        return ;
    }//无解
    bool ok = false;
    while(1)
    {
        if(l >= m)    break;
        int i, pos = 0;
        ok = false;//判断变量ok，重要！
        for(i = 0; i < n; i++)
        {
            if(seg[i].L <= l)
            {
                if(seg[i].R > r)
                {
                    pos = i;
                    r = seg[pos].R;
                    ok = true;
                }//寻找起点在l的最长区间
            }
        }
        if(ok)
        {
            l = r;
            ans[cnt].L = seg[pos].L;
            ans[cnt++].R = seg[pos].R;
        }//更新起点l
        else break;
    }
    if(ok)
    {
        printf("%d
", cnt);
        for(int i = 0; i < cnt; i++)
            printf("%d %d
", ans[i].L, ans[i].R);
    }
    else printf("0
");
}
int main()
{
    int T;
    scanf("%d", &T);
    for(int kase = 0; kase < T; kase++)
    {
        if(kase) printf("
");
        memset(seg, 0, sizeof(seg));
        memset(ans, 0, sizeof(ans));
        int m; scanf("%d", &m);
        int i = 0, L, R;
        while(scanf("%d%d", &L, &R))
        {
            if(!L && !R)  break;
            seg[i].L = L; seg[i].R = R;
            i++;
        }
        sort(seg, seg+i);//排序
        solve(i, m);
    }
    return 0;
}