杭电acm 1009 FatMouse' Trade 比较典型的贪心问题

本题可以作为贪心问题的水题，我刚开始时，还掉以轻心，后第一遍wa后，我才发现这道题并不那么简单。

首先，分析该题：输入后应先进行排序，按照j/f来进行排序。后再尽可能选取j/f比较大，直到m==0.

但是，问题到这并没有结束，请你接下来试试这几组数据：

0 11 01.0001 00.0005 410000 52000 2100 0300 010400.000

是不是wa了；

这才是本题问题所在。

后经过修改，源码如下：

#include <iostream>
#include <cstdio>

using namespace std;

typedef struct{
            double ans;
            int j;
            int f;
}ROOM;

ROOM room[1010];
int m,n;

int main(){
            while(scanf("%d%d",&m,&n)!=EOF&&(m!=-1||n!=-1)){
                        for(int i=0;i<n;i++){
                                    scanf("%d %d",&room[i].j,&room[i].f);
                                    room[i].ans =1.0* room[i].j/room[i].f;
                        }
                        for(int i=0;i<n-1;i++){
                                    for(int j=i+1;j<n;j++){
                                                if(room[j].ans>room[i].ans){
                                                            ROOM temp = room[i];
                                                            room[i] = room[j];
                                                            room[j] = temp;
                                                }
                                    }
                        }
                        int i=0;
                        double count=0;
                        for(int i=0;i<n;i++){
                                    if(room[i].f==0)        {
                                                            count+=room[i].j;
                                                            room[i].ans = 0;
                                                            room[i].f=0;
                                                            room[i].j=0;
                                    }
                        }
                        while(m){
                                    if(room[i].f<=m){
                                                count += room[i].j;
                                                m -= room[i].f;
                                                i++;
                                    }
                                    else{
                                                count += room[i].ans*m;
                                                break;
                                    }
                        }
                        if(n==0)    printf("0.000\n");
                        else printf("%.3f\n",count);
            }
            return 0;
}