HDU 4004 The Frog's Games 二分 贪心

戳这里：HDU 4004

//思路：二分经典入门题...贪心判方案是否可行

 1 #include "bits/stdc++.h"
 2 using namespace std;
 3 int L, n, m;
 4 int pos[500010], dis[500010];
 5
 6 bool Cant(int Dis_Jump)
 7 {
 8     int i, Dis_Sum = 0, Count = 0;
 9     for(i = 1; i <= n + 1; ++i) {
10         if(dis[i] > Dis_Jump) {
11             return 1;
12         }
13         Dis_Sum += dis[i];
14         if(Dis_Sum > Dis_Jump) {
15             ++Count;
16             Dis_Sum = dis[i];
17         }
18     }
19     ++Count;
20     if(Count > m)
21         return 1;
22     return 0;
23 }
24
25 int main()
26 {
27     while(scanf("%d%d%d", &L, &n, &m) != EOF) {
28         int i;
29         for(i = 1; i <= n; ++i) {
30             scanf("%d", &pos[i]);
31         }
32         sort(pos + 1, pos + 1 + n);
33         for(i = 1; i <= n; ++i) {
34             dis[i] = pos[i] - pos[i - 1];
35         }
36         dis[i] = L - pos[i - 1];
37 //        for(i = 1; i <= n + 1; ++i) {
38 //            printf(" ee i == %d dis == %d\n", i, dis[i]);
39 //        }
40
41         int l = 0, r = L, mid;
42         while(l <= r) {
43             mid = (l + r) >> 1;
44             if(Cant(mid)) {
45                 l = mid + 1;
46             }
47             else {
48                 r = mid - 1;
49             }
50         }
51         printf("%d\n", l);
52     }
53 }

HDU 4004 The Frog's Games 二分 贪心