poj 3262 Protecting the Flowers(贪心)

Protecting the Flowers Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 3528   Accepted: 1442 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 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 USACO 2007 January Silver

题意：

有n头牛在糟蹋庄稼。把第i头牛牵回家需要ti分钟。第i头牛每分钟会摧毁di的庄稼。每次只能牵一头牛走。问怎么牵使损失最少。

思路：

考虑a,b两牛。先牵a牛和b牛的损失分别为。2*d[b]*t[a]，2*d[a]*t[b]。设先牵a更优。2*d[b]*t[a]<2*d[a]*t[b].

所以根据优先级排序然后依次牵就是最优的选择。

详细见代码：

#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
//#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const double PI=acos(-1.0);
const int maxn=100010;
typedef __int64 ll;
struct node
{
    int t,d;
} cow[maxn];
bool cmp(node a,node b)
{
    return a.t*b.d<b.t*a.d;
}
int main()
{
    int n,i;
    ll ans,ti;

    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
            scanf("%d%d",&cow[i].t,&cow[i].d);
        sort(cow,cow+n,cmp);
        ans=ti=0;
        for(i=0;i<n;i++)
        {
            ans+=ti*cow[i].d;
            ti+=2*cow[i].t;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}