【題目描述】
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
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.
【分析】
這道題是貪心策略，我們可以用一個C結構儲存每頭奶牛的T、D。
貪心策略就是找出C[i].T/C[i].D的最小值，這可以用快速排序來實現。
大家知道除法運算時很慢的，為了提高速度，我們可以把 快速排序的換元函數中的 (C[i].T/C[i].D)<(C[j].T/C[j].D) 交叉相乘，改為 (C[i].T*C[j].D)<(C[j].T*C[i].D)。
快速排序後，位於數組首位的奶牛不毀壞花，所以從i=2 to N枚舉每頭牛破壞的花的個數，輸出即可。由於結果可能很大，最好用long long。
【我的代碼】
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
int N;
class COW
{
public:
int speed;
int eat;
}C[100001];
int cmp(const void *a,const void *b)
{
class COW *c=(class COW *)a;
class COW *d=(class COW *)b;
int A=c->speed*d->eat;
int B=d->speed*c->eat;
if(A<B)
return -1;
else
return 1;
}
void init()
{
scanf("%d\n",&N);
for (int i=1;i<=N;i++)
scanf("%d %d\n",&C[i].speed,&C[i].eat);
qsort(C+1,N,sizeof(C[0]),cmp);
long long T=0,Des=0;
for (int i=1;i<=N;i++)
{
Des+=(T*C[i].eat);
T+=2*C[i].speed;
}
cout<<Des<<endl;
}
int main()
{
freopen("flowers.in","r",stdin);
freopen("flowers.out","w",stdout);
init();
return 0;
}
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
- Line 1: A single integer that is the minimum number of destroyed flowers
6 3 1 2 5 2 3 3 2 4 1 1 6Sample Output
86Output Details
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.
【分析】
這道題是貪心策略，我們可以用一個C結構儲存每頭奶牛的T、D。
貪心策略就是找出C[i].T/C[i].D的最小值，這可以用快速排序來實現。
大家知道除法運算時很慢的，為了提高速度，我們可以把 快速排序的換元函數中的 (C[i].T/C[i].D)<(C[j].T/C[j].D) 交叉相乘，改為 (C[i].T*C[j].D)<(C[j].T*C[i].D)。
快速排序後，位於數組首位的奶牛不毀壞花，所以從i=2 to N枚舉每頭牛破壞的花的個數，輸出即可。由於結果可能很大，最好用long long。
【我的代碼】
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
int N;
class COW
{
public:
int speed;
int eat;
}C[100001];
int cmp(const void *a,const void *b)
{
class COW *c=(class COW *)a;
class COW *d=(class COW *)b;
int A=c->speed*d->eat;
int B=d->speed*c->eat;
if(A<B)
return -1;
else
return 1;
}
void init()
{
scanf("%d\n",&N);
for (int i=1;i<=N;i++)
scanf("%d %d\n",&C[i].speed,&C[i].eat);
qsort(C+1,N,sizeof(C[0]),cmp);
long long T=0,Des=0;
for (int i=1;i<=N;i++)
{
Des+=(T*C[i].eat);
T+=2*C[i].speed;
}
cout<<Des<<endl;
}
int main()
{
freopen("flowers.in","r",stdin);
freopen("flowers.out","w",stdout);
init();
return 0;
}
