标签:tor class -- ace freopen opera 比较 can clu
http://poj.org/problem?id=3262
这道题和蝎子那道题是相同贪心思路
http://www.cnblogs.com/oscar-cnblogs/p/6329703.html
//贪心方式一:列出函数关系式比较复杂
//方式二:像蝎子那道题 先讨论两只 奶牛怎样送是最优的
/*
a1 b1
a2 b2
先送1 ans = 2*a1*b2
先送2 ans = 2*b1*a2
那么如果先送1 更优说明 2*a1*b2 < 2*b1*a2
--->> a1 / b1 < a2 / b2这样就得出比较方式了
*/
1 #include <iostream> 2 #include <stdio.h> 3 #include <algorithm> 4 using namespace std; 5 6 7 int n; 8 struct Cow 9 { 10 double t, d; 11 bool operator < (Cow a) const 12 { 13 return t / d < a.t / a.d; 14 } 15 }cow[100007]; 16 int main() 17 { 18 freopen("in.txt", "r", stdin); 19 while(~scanf("%d", &n)) 20 { 21 long long ans = 0, sum = 0; 22 for (int i = 0; i < n; i++) 23 { 24 scanf("%lf%lf", &cow[i].t, &cow[i].d); 25 sum += cow[i].d; 26 } 27 sort(cow, cow+n); 28 for (int i = 0; i < n; i++) 29 { 30 sum -= cow[i].d; 31 ans += sum*cow[i].t*2; 32 //printf("%.0lf %.0lf\n", cow[i].t, cow[i].d); 33 } 34 printf("%lld\n", ans); 35 } 36 return 0; 37 }
POJ 3262 Protecting the flowers
标签:tor class -- ace freopen opera 比较 can clu
原文地址:http://www.cnblogs.com/oscar-cnblogs/p/6376273.html