UVA - 1346 Songs 贪心

题目大意：给出n首歌曲，每首歌曲有相对应的ID，长度L和频率F，现在要讲这n首歌曲刻录在一张唱片上，要求sum的值达到最小，问该如何刻录

sum = ∑i=1n\displaystyle \sum^{{n}}_{{i=1}}fs(i)∑j=1s(i)\displaystyle \sum^{{s(i)}}_{{j=1}}ls(j)

fs(i)表示第i首歌的频率，ls(j)表第j首歌的长度

解题思路：要让sum最小的最，就要让长度小的尽量排前，频率大的尽量排前

k = 长度/频率，按照没首歌的k值排序即可

证明如下，假设：

设：sum1 = A(i) + A(i+1) = (fs(i) + fs(i+1)) * (ls(1) + ls(2) + … +ls(i-1)) + fs(i) * ls(i) + fs(i+1) * ls(i) + fs(i+1) * ls(i+1)

交换第i首和第i+1首的次序

sum2 = A(i+1) + A(i) = (fs(i) + fs(i+1)) * (ls(1) + ls(2) + … +ls(i-1)) + fs(i) * ls(i) + fs(i+1) * ls(i+1) + fs(1) * ls(i+1)

因为sum1>sum2，所以fs(i+1) * ls(i) > fs(i) * ls(i+1),可得，fs(i+1) / ls(i+1) > fs(i) / ls(i)

[code]#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = (1 << 16) + 10;

struct tap{
    int name;
    double k;
    bool operator <(const tap &t) const {
        return k < t.k;
    }
}T[maxn]; 

int main() {
    int N;
    while(scanf("%d",&N) == 1) {
        int n;
        double f, l, k;
        for(int i = 0; i < N; i++)   {
            scanf("%d%lf%lf",&n, &l, &f);
            k = l / f;  
            T[i].name = n;
            T[i].k = k; 
        }
        sort(T,T+N);
        int num;
        scanf("%d",&num);
        printf("%d\n",T[num-1].name);

    }
    return 0;
}