标签:tac 分析 eof algorithm tor eps mat amp print
题意:把c个牛分进n个摊位,摊位位置已知,所有摊位分布在0 <= xi <= 1,000,000,000,问两头牛间最小距离的最大值。
分析:找所有最小距离取个最大的。所以二分找这个最小的距离,这个最大的最小距离是二分的分界线。贪心来判断当前的最小距离是否能安排下所有的牛。
#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) typedef long long ll; typedef unsigned long long llu; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const ll LL_INF = 0x3f3f3f3f3f3f3f3f; const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos(-1.0); const double eps = 1e-8; const int MAXN = 100000 + 10; const int MAXT = 10000 + 10; using namespace std; int a[MAXN]; int n, c; bool judge(int x){ int cnt = 1; int st = a[0]; for(int i = 1; i < n; ++i){ if(a[i] - st >= x){ ++cnt; st = a[i]; } if(cnt == c) return true; } return false; } int solve(){ int l = 1, r = 1e9; while(l < r){ int mid = l + (r - l + 1) / 2; if(judge(mid)) l = mid; else r = mid - 1; } return l; } int main(){ while(scanf("%d%d", &n, &c) == 2){ memset(a, 0, sizeof a); for(int i = 0; i < n; ++i){ scanf("%d", &a[i]); } sort(a, a + n); int ans = solve(); printf("%d\n", ans); } return 0; }
POJ - 2456 Aggressive cows(二分+贪心)
标签:tac 分析 eof algorithm tor eps mat amp print
原文地址:http://www.cnblogs.com/tyty-Somnuspoppy/p/6271022.html