UVa-10020 Minimal coverage **

 1 /*
 2  * Uva-10020 最少区间覆盖-贪心
 3  *
 4  * 很直接的贪心。。 详见刘汝佳 《奥赛入门经典》
 5  *
 6  *    测试数据： http://www.algorithmist.com/index.php/UVa_10020#Input
 7  *
 8  */
 9 
10 #include <cstdio>
11 #include <algorithm>
12 using namespace std;
13 
14 const int maxN = 100000 + 5;
15 int caseNum, m, intervalNum, ansList[maxN];
16 
17 struct SData{
18     int x, y;
19 };
20 SData interval[maxN];
21 
22 bool cmp(const SData lhs, const SData rhs){
23     return lhs.x < rhs.x;
24 }
25 
26 int main(){
27     scanf("%d", &caseNum);
28     while(caseNum--){
29         scanf("%d", &m);
30 
31         intervalNum = 0;
32         int tmpX, tmpY;
33         while(scanf("%d%d", &tmpX, &tmpY)){
34             if(tmpX == 0 && tmpY == 0)
35                 break;
36 
37             interval[intervalNum].x = tmpX;
38             interval[intervalNum++].y = tmpY;
39         }
40 
41         sort(interval, interval + intervalNum, cmp);
42 
43         int cover = 0, cur = 0, nextCover, nextCoverNum, i, ans = 0;
44         bool covered = 0;    //注意初始化！！在这里ＷＡ了一次
45         while(cur < intervalNum){
46             nextCover = cover;
47             i = cur;
48             while(i < intervalNum && interval[i].x <= cover){
49                 if(interval[i].y > nextCover){
50                     nextCover = interval[i].y;
51                     nextCoverNum = i;
52                 }
53                 i++;
54             }
55             if(i == cur) break;
56 
57             cover = nextCover;
58             ansList[ans++] = nextCoverNum;
59             cur = i;
60             if(cover >= m){
61                 covered = 1;
62                 break;
63             }
64         }
65         if(covered){
66             printf("%d\n", ans);
67             for(int j=0; j<ans; j++){
68                 printf("%d %d\n", interval[ansList[j]].x, interval[ansList[j]].y);
69             }
70         }
71         else printf("0\n");
72         if(caseNum > 0) printf("\n");
73 
74     }
75 
76 
77     return 0;
78 }

　　　　

 1 /*
 2 另外，针对区间端点为整数的情形， 可以直接用数组存，不用排序，O(n)时间内解决。
 3 即interval[i] = j， 表示以左端点为i的区间的右端点最大值为j。。
 4 */
 5 #include <cstdio>
 6 #include <cstring>
 7 using namespace std;
 8 
 9 const int maxn = 50000 + 10;
10 int m, interval[maxn];
11 
12 int main(){
13     memset(interval, -1, sizeof(interval));
14 
15     scanf("%d", &m);
16     int tmpa, tmpb;
17     while(scanf("%d%d", &tmpa, &tmpb)){
18         if(tmpa==0 && tmpb==0) break;
19         if(tmpb < 0 || tmpa >= m) continue;
20         if(tmpa < 0) tmpa = 0;
21         if(interval[tmpa] < tmpb)
22             interval[tmpa] = tmpb;
23     }
24 
25     int cover=0, cur=0, i=0, imax = 0, num = 0;
26     bool flag = 0;
27     while(i<m){
28         flag = 0;
29         for(; i<=cover; i++){
30             if(interval[i] > imax){
31                 imax = interval[i];
32                 cur = i;
33             }
34         }
35         if(imax == cover) break;
36         
37         cover = imax;
38         num++;
39         if(cover >= m){
40             flag = 1;
41             break;
42         }
43     }
44     
45     if(flag) printf("%d\n", num);
46     else printf("No solution\n");
47 
48     return 0;
49 }