【BZOJ】1634: [Usaco2007 Jan]Protecting the Flowers 护花(贪心)
http://www.lydsy.com/JudgeOnline/problem.php?id=1634
贪心。。
我们发现,两个相邻的牛(a和b)哪个先走对其它的牛无影响,但是可以通过 a的破坏花×b的时间 和 b的破坏花×a的时间 可以判断哪个先走。
那么可以应用到所有牛上。。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; } const int N=100005;
int n;
struct nod { int a, b; }cow[N];
bool cmp(const nod &a, const nod &b) { return b.a*a.b>a.a*b.b; }
int main() {
read(n);
for1(i, 1, n) read(cow[i].a), read(cow[i].b);
long long ans=0, sum=0;
sort(cow+1, cow+1+n, cmp);
for1(i, 1, n) sum+=cow[i].b;
for1(i, 1, n) {
sum-=cow[i].b;
ans+=cow[i].a*sum;
}
printf("%lld", ans<<1);
return 0;
}
Description
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 cows were in his garden eating his beautiful flowers. Wanting to minimize the subsequent damage, FJ decided to take immediate action and transport the cows back to their barn. Each cow i is at a location that is Ti minutes (1 <= Ti <= 2,000,000) away from the 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 the barn. Moving cow i to the barn requires 2*Ti minutes (Ti to get there and Ti to return). 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
第1行输入N,之后N行每行输入两个整数Ti和Di.
Output
* Line 1: A single integer that is the minimum number of destroyed flowers
一个整数,表示最小数量的花朵被吞食.
Sample Input
3 1
2 5
2 3
3 2
4 1
1 6
Sample Output
HINT
约翰用6,2,3,4,1,5的顺序来运送他的奶牛.
Source
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