Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15649 Accepted Submission(s): 5747
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 .
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
Source
题目意思：
一个人去抢劫n个银行，初始给一个概率p，每个银行有能抢劫的价值和被抓住的概率，当抢劫银行被抓住的总概率小于p的时候认为这个人是安全的，求在安全状态下能抢劫到最多得价值。
一个人去抢劫n个银行，初始给一个概率p，每个银行有能抢劫的价值和被抓住的概率，当抢劫银行被抓住的总概率小于p的时候认为这个人是安全的，求在安全状态下能抢劫到最多得价值。
思路：
若以被抓概率为体积、价值为价值，概率是浮点数无法作为数组下标，不可取。
若以被抓概率为体积、价值为价值，概率是浮点数无法作为数组下标，不可取。
若以价值为体积、被抓概率为价值，而背包的定义是体积一定时获得最多得价值和使得被抓概率尽可能小相反，不可取。
若以价值为体积、不被抓概率为价值，使得不被抓概率尽可能高，可取。
代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
using namespace std; #define N 105 int max(int x,int y){return x>y?x:y;}
int min(int x,int y){return x<y?x:y;}
int abs(int x,int y){return x<?-x:x;} int n;
int v[N];
double val[N];
double dp[N*N];
int V; main()
{
int i, j, k;
int t;
cin>>t;
while(t--){
double vv;
scanf("%lf %d",&vv,&n);
V=;
for(i=;i<=n;i++) {
scanf("%d %lf",&v[i],&val[i]);
val[i]=1.0-val[i];
V+=v[i];
}
memset(dp,,sizeof(dp));
dp[]=;
for(i=;i<=n;i++){
for(j=V;j>=v[i];j--)
dp[j]=max(dp[j],dp[j-v[i]]*val[i]);
}
for(i=V;i>=;i--){
if(dp[i]>=-vv) break;
}
if(i<) printf("0\n");
else
printf("%d\n",i);
}
}