HDU——2955Robberies（小数背包）

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19862    Accepted Submission(s): 7344

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

 

这题很久以前做过，当时不是很懂，看着别人的思路才写出来的，而且不是很懂思路中的转换，所以这次重新做一遍。

思路：题目中给定价值和被抓几率，但是被抓几率不可以用乘积来组合计算，举个例子，比如第一个银行3%被抓几率，第二个5%被抓几率，那么乘起来会变成0.15%，抢的越多，被抓几率却越小了，显然不对，因此要转换成不被抓几率，上述例子则变为第一家97%不被抓，第二家95%不被抓，乘起来就是92.15%，抢的越多，不被抓的几率越来越小即被抓几率越来越大，这样才是符合常理的。

那么背包体积应该是什么呢？先看最普通01背包，用数个cost来填充V，使得value之和尽量大，那么这题就应该是用数个money填充总money，使得不被抓几率尽量大。那转移方程就是dp[j]=max（dp[j]，dp[j-w]*c），这里和01背包的区别就是从+改成了*。然后得到dp数组是0~V情况下的不被抓几率的最优（大）值，这根答案有什么关系？逆序枚举每一种情况，若此情况下的dp值即不被抓几率大于等于题目中所给的不被抓几率，那就输出，逆序着从大到小枚举保证了找到的一个解是最优解。

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x) memset(x,0,sizeof(x))
#define MMINF(x) memset(x,INF,sizeof(x))
typedef long long LL;
const double PI=acos(-1.0);
const int N=110;
struct info
{
	int w;
	double c;
};
info bank[N];
double dp[N*100];
bool cmp(const info &a,const info &b)
{
	return a.w<b.w;
}
int main(void)
{
	int tcase,i,j,V,n;
	double k;
	scanf("%d",&tcase);
	while (tcase--)
	{
		V=0;
		scanf("%lf%d",&k,&n);
		k=1.0-k;
		for (i=0; i<n; i++)
		{
			scanf("%d%lf",&bank[i].w,&bank[i].c);
			bank[i].c=1.0-bank[i].c;
			V+=bank[i].w;
		}
		sort(bank,bank+n,cmp);
		for (i=0; i<N*100; i++)
			dp[i]=0;
		dp[0]=1;
		for (i=0; i<n; i++)
		{
			for (j=V; j>=bank[i].w; j--)
			{
				dp[j]=max(dp[j],dp[j-bank[i].w]*bank[i].c);
			}
		}
		for (i=V; i>=0; i--)
		{
			if(dp[i]>=k)
			{
				printf("%d\n",i);
				break;
			}
		}
	}
	return 0;
}