标签：open logs ant strong event 二分 vol std any
题目链接：http://poj.org/problem?id=3258
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 15753 | Accepted: 6649 |
Description
Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L).
To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river.
Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N).
FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks.
Input
Output
Sample Input
25 5 2 2 14 11 21 17
Sample Output
4
Hint
Source
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 #define ms(a,b) memset((a),(b),sizeof((a))) 13 using namespace std; 14 typedef long long LL; 15 const double EPS = 1e-8; 16 const int INF = 2e9; 17 const LL LNF = 2e18; 18 const int MAXN = 1e5; 19 20 int L, N, M; 21 int a[MAXN]; 22 23 bool test(int mid) 24 { 25 int cnt = 0, pre = 1; 26 for(int i = 2; i<=N; i++) 27 { 28 if(a[i]-a[pre]<mid) cnt++; 29 else pre = i; 30 } 31 return cnt<=M; 32 } 33 34 int main() 35 { 36 while(scanf("%d%d%d",&L, &N, &M)!=EOF) 37 { 38 for(int i = 1; i<=N; i++) 39 scanf("%d",&a[i]); 40 41 a[++N] = 0; a[++N] = L; 42 sort(a+1, a+1+N); 43 44 int l = 1, r = L; 45 while(l<=r) 46 { 47 int mid = (l+r)>>1; 48 if(test(mid)) 49 l = mid + 1; 50 else 51 r = mid - 1; 52 printf("%d %d\n", l, r); 53 } 54 printf("%d\n", r); 55 } 56 }
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 #define ms(a,b) memset((a),(b),sizeof((a))) 13 using namespace std; 14 typedef long long LL; 15 const double EPS = 1e-8; 16 const int INF = 2e9; 17 const LL LNF = 2e18; 18 const int MAXN = 1e5; 19 20 int L, N, M; 21 int a[MAXN], tmp[MAXN]; 22 23 bool test(int mid) 24 { 25 memcpy(tmp, a, sizeof(tmp)); //!!!!! 26 int cnt = 0; 27 for(int i = 1; i<=N; i++) 28 if(tmp[i]<mid) 29 cnt++, tmp[i+1] += tmp[i]; 30 return cnt<=M; 31 } 32 33 int main() 34 { 35 while(scanf("%d%d%d",&L, &N, &M)!=EOF) 36 { 37 for(int i = 1; i<=N; i++) 38 scanf("%d",&a[i]); 39 40 a[++N] = 0; a[++N] = L; 41 sort(a+1, a+1+N); 42 for(int i = 1; i<N; i++) 43 a[i] = a[i+1]-a[i]; 44 N--; 45 46 int l = 1, r = L; 47 while(l<=r) 48 { 49 int mid = (l+r)>>1; 50 if(test(mid)) 51 l = mid + 1; 52 else 53 r = mid - 1; 54 } 55 printf("%d\n", r); 56 } 57 }
标签：open logs ant strong event 二分 vol std any
原文地址：http://www.cnblogs.com/DOLFAMINGO/p/7545779.html
