Protecting the Flowers
Farmer John went to cut some wood and left N (2 ≤ N ≤ 100,000) cows eating the grass, as usual. When he returned, he found to his horror that the cluster of cows was in his garden eating his beautiful flowers. Wanting to minimize the subsequent damage, FJ decided to take immediate action and transport each cow back to its own barn.
Each cow i is at a location that is Ti minutes (1 ≤ Ti ≤ 2,000,000) away from its own barn. Furthermore, while waiting for transport, she destroys Di (1 ≤ Di ≤ 100) flowers per minute. No matter how hard he tries, FJ can only transport one cow at a time back to her barn. Moving cow i to its barn requires 2 × Ti minutes (Ti to get there and Ti to return). FJ starts at the flower patch, transports the cow to its barn, and then walks back to the flowers, taking no extra time to get to the next cow that needs transport.
Write a program to determine the order in which FJ should pick up the cows so that the total number of flowers destroyed is minimized.
Input
Line 1: A single integer N
Lines 2..N+1: Each line contains two space-separated integers, Ti and Di, that describe a single cow's characteristics
Output
Line 1: A single integer that is the minimum number of destroyed flowers
Sample Input
6
3 1
2 5
2 3
3 2
4 1
1 6
Sample Output
86
Hint
FJ returns the cows in the following order: 6, 2, 3, 4, 1, 5. While he is transporting cow 6 to the barn, the others destroy 24 flowers; next he will take cow 2, losing 28 more of his beautiful flora. For the cows 3, 4, 1 he loses 16, 12, and 6 flowers respectively. When he picks cow 5 there are no more cows damaging the flowers, so the loss for that cow is zero. The total flowers lost this way is 24 + 28 + 16 + 12 + 6 = 86.
Source
题目类型：贪心
算法分析：对于序列中相邻的两个奶牛A和B，若A在B之前，则加入后的总费用Asum为(sumd - Ad) * 2At + (sumd - Ad - Bd) * 2Bt，若B在前，则总费用Bsum为(sumd - Bd) * 2Bt + (sumd - Bd - Ad) * 2At，其中sumd表示当前还没有被运输的奶牛单位时间的总损害量。若A要在B的前面，则需要满足Asum < Bsum，则将上式代入后化简得Bd / Bt < Ad / At，即单位时间损害量大的要先被拿走
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 |
/************************************************* Author :supermaker Created Time :2016/4/14 8:42:28 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e5 + 6666; struct Node { LL t, d; double k; bool operator < (const Node &a) const { return k > a.k; } }; Node aa[maxn]; int main() { //CFF; //CPPFF; int n; while (scanf ("%d", &n) != EOF) { LL sumd = 0; for (int i = 1; i <= n; i++) { scanf ("%lld%lld", &aa[i].t, &aa[i].d); aa[i].k = aa[i].d * 1.0 / aa[i].t; sumd += aa[i].d; } sort (aa + 1, aa + 1 + n); LL res = 0; for (int i = 1; i <= n; i++) { res += (sumd - aa[i].d) * 2 * aa[i].t; sumd -= aa[i].d; } printf ("%lld\n", res); } return 0; } |
- « 上一篇：poj3040
- poj2718：下一篇 »