【区间覆盖问题】uva 10020 - Minimal coverage

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 = ;
 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 = , l = , r = ;//起点l，起点l最长区间的右端点r
     if(seg[].L > l)
     {
         printf("0\n");
         return ;
     }//无解
     bool ok = false;
     while()
     {
         if(l >= m)    break;
         int i, pos = ;
         ok = false;//判断变量ok，重要！
         for(i = ; 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\n", cnt);
         for(int i = ; i < cnt; i++)
             printf("%d %d\n", ans[i].L, ans[i].R);
     }
     else printf("0\n");
 }
 int main()
 {
     int T;
     scanf("%d", &T);
     for(int kase = ; kase < T; kase++)
     {
         if(kase) printf("\n");
         memset(seg, , sizeof(seg));
         memset(ans, , sizeof(ans));
         int m; scanf("%d", &m);
         int i = , 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 ;
 }