求教hdoj上一个题目
2955
Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3594 Accepted Submission(s): 1344
Problem Description
The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative business of bank robbery only for a short while, before retiring to a comfortable job at a university.
For a few months now, Roy has been assessing the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible.
His mother, Ola, has decided upon a tolerable probability of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.
Input
The first line of input gives T, the number of cases. For each scenario, the first line of input gives a floating point number P, the probability Roy needs to be below, and an integer N, the number of banks he has plans for. Then follow N lines, where line j gives an integer Mj and a floating point number Pj .
Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .
Output
For each test case, output a line with the maximum number of millions he can expect to get while the probability of getting caught is less than the limit set.
Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.
Sample Input
3 0.04 3 1 0.02 2 0.03 3 0.05 0.06 3 2 0.03 2 0.03 3 0.05 0.10 3 1 0.03 2 0.02 3 0.05
Sample Output
2 4 6[code=C/C++][/code]
#include<iostream>
using namespace std;
struct node
{
float w;
int q;
}s[105];
int m[105];
void sort(int n)
{
int i,j,l,t;
float k;
for(i=1;i<=n;i++)
{
for(l=i,j=i+1;j<=n;j++)
if((s[l].w/s[l].q)<(s[j].w/s[j].q))l=j;
if(l!=i)
{
t=s[i].q;s[i].q=s[l].q;s[l].q=t;
k=s[i].w;s[i].w=s[l].w;s[l].w=k;
}
}
}
int main()
{
int i,j,k,n,max;
float d,x;
cin>>k;
while(k--)
{
cin>>d>>n;
for(i=1;i<=n;i++)
cin>>s[i].q>>s[i].w;
sort(n);
for(max=0,i=1;i<=n;i++)
{
for(x=0,j=i;j<=n;j++)
if(x+s[j].w<=d){m[i]+=s[j].q;x+=s[j].w;}
max=max<m[i]?m[i]:max;
}
cout<<max<<endl;
memset(m,0,sizeof(m));
}
return 0;
}
我的主要思路：先根据效率（危险度除以钱）降序排序，再一个个求在不超过给定限度的最大收获，最后求的最大值再输出
我提交的结果是WA.。求各路大牛揪出我算法里头的错误。。
[解决办法]
楼主，这样排序是不行的，你看下这组数据：
0.08 4
2 0.01
4 0.02
8 0.04
16 0.07
正确答案应该为 18吧
