UVA 1346 - Songs (贪心)

来源：互联网发布：佐伊技能数据编辑：程序博客网时间：2024/05/16 14:24

题意：n张唱片，每张都有id，长度f 和频率a，要求题目中公式值最小。
求一个排序，输出该排序下第m张CD的id。

解析：
证明：
sum = b[1]+...+b[n] , 令b[i]= f[i]*(a[1]+..+a[i]);

假设以求出最小的sum 得到了最优序列那么

sum1 = b[i]+b[i+1] = f[i]*(a[1]+...a[i]) + f[i+1]*(a[1]+...+a[i+1])
= (a[1]+...+a[i-1])*(f[i]+f[i+1]) + f[i+1]*a[i+1]+ f[i+1]*a[i] + f[i]*a[i];

现在交换 i 和 i+1
sum2 = (a[1]+...+a[i-1])*(f[i]+f[i+1]) + f[i+1]*a[i+1]+ f[i]*a[i+1] + f[i]*a[i];
ans = sum1 - sum2 = f[i+1]*a[i] - f[i]*a[i+1]

if ans < 0
swap (i,i+1);

即 f[i+1]*a[i] < f[i]*a[i+1]
a[i] / f[i] < a[i+1] / f[i+1]
swap (i,i+1);

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>using namespace std;typedef long long ll;const int INF = 0x3f3f3f3f;struct Song {int id;double f, l;}song[1 << 17];bool cmp(Song a,Song b) {return (a.l * b.f) < (b.l * a.f);}int main() {int n ,m;while(scanf("%d",&n) != EOF) {for(int i = 1; i <= n; i++) {scanf("%d%lf%lf",&song[i].id,&song[i].l,&song[i].f);}sort(song+1,song+n+1,cmp);scanf("%d",&m);printf("%d\n",song[m].id);}return 0;}

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