POJ - 3258 River Hopscotch(二分 最大化最小值)

大致题意:

　　坐标 从  [0-L], 上有 N 个点 , 现在需要移去 M 个点.求任意两点(包括0,L)点的距离中最小值的最大值

思路:枚举答案,若移去M个点不能达成目标,则使答案变小.

while(l <= r){
     ll mid = l + (r-l)/2;
     // 是否可能使得所有石头之间的距离不小于 mid
     if(check(mid)){
          l = mid+1;
     }else{
          r = mid-1;
     }
}
printf("%lld\n",l-1);    

 1 bool check(ll mid){
 2     ll cnt = 0;
 3     ll pre = 0;
 4     for(ll i = 1 ; i <= N ; i ++){
 5         if(D[i]-D[pre]>=mid){　　
 6             pre = i;
 7         }else{　　　　//两块之间距离大于要求距离的话,移去该石头
 8             cnt++;
 9         }
10         if(cnt > M)    return false;
11     }
12     return true;
13 }

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4
 5 #include<string>
 6 #include<vector>
 7 #include<set>
 8 #include<map>
 9 #include<queue>
10 #include<math.h>
11
12 #include<algorithm>
13 #include<iostream>
14
15 const int INF = 0x7f7f7f7f;
16 using namespace std;
17 typedef long long ll;
18
19 /*
20     1<=L<=100000000
21     N:rocks
22     M:remove rocks
23     Di:
24     L N M:
25 */
26
27 ll D[50005];
28 int L,N,M;
29
30 bool check(ll mid){
31     ll cnt = 0;
32     ll pre = 0;
33     for(ll i = 1 ; i <= N ; i ++){
34         if(D[i]-D[pre]>=mid){
35             pre = i;
36         }else{
37             cnt++;
38         }
39         if(cnt > M)    return false;
40     }
41     return true;
42 }
43
44 int main(){
45     while(scanf("%d%d%d",&L,&N,&M) != EOF){
46         D[0] = 0;
47         for(int i = 1 ; i <= N ; i ++)    scanf("%lld",&D[i]);
48         D[++N] = L;
49         sort(D,D+N);
50         ll l = 0 , r = 1000000000;
51         while(l <= r){
52             ll mid = l + (r-l)/2;
53             // 是否可能使得所有石头之间的距离不小于 mid
54             if(check(mid)){
55                 l = mid+1;
56             }else{
57                 r = mid-1;
58             }
59         }
60         printf("%lld\n",l-1);
61     }
62
63     return 0;
64 }