DP专题训练之HDU 2955 Robberies

打算专题训练下DP，做一道帖一道吧~~现在的代码风格完全变了~~大概是懒了。所以。将就着看吧~哈哈

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.

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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

题目大意就是：Roy抢劫银行，每家银行都有一定的金额和被抓到的概率，在已知Roy被抓的概率的情况下，求Roy被抓住情况下，可以抢到的最多的钱。

这个题目很坑的点，就是精度不止两位，但是样例却又是。。。。。。唉~

特殊数据就是Roy被抓到的概率接近于0的时候，Roy抢到的钱就是所有银行的金额的和~

简单01背包而已，注意精度问题~~

//Asimple
//#include <bits/stdc++.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cctype>
#include <cstdlib>
#include <stack>
#include <cmath>
#include <set>
#include <map>
#include <string>
#include <queue>
#include <limits.h>
#include <time.h>
#define INF 0xfffffff
#define mod 1000000
#define swap(a,b,t) t = a, a = b, b = t
#define CLS(a, v) memset(a, v, sizeof(a))
#define debug(a)  cout << #a << " = "  << a <<endl
#define abs(x) x<0?-x:x
#define srd(a) scanf("%d", &a)
#define src(a) scanf("%c", &a)
#define srs(a) scanf("%s", a)
#define srdd(a,b) scanf("%d %d",&a, &b)
#define srddd(a,b,c) scanf("%d %d %d",&a, &b, &c)
#define prd(a) printf("%d\n", a)
#define prdd(a,b) printf("%d %d\n",a, b)
#define prs(a) printf("%s\n", a)
#define prc(a) printf("%c", a)
using namespace std;
typedef long long ll;
const int maxn = 105;
int n, m, num, T, k, len, ans, sum;
double dp[10005];
double pp[maxn], p;
int mon[maxn];
void input() {
    srd(T);
    while( T -- ) {
        scanf("%lf %d", &p, &n);
        sum = 0;
        for(int i=0; i<n; i++) {
            scanf("%d %lf",&mon[i],&pp[i]);
            sum += mon[i];
        }
        CLS(dp, 0);
        dp[0] = 1;//金额为0时安全 
        if( p <= 1-1e-8 ) {//有被抓的概率
            int i; 
            for(i=0; i<n; i++) {
                for(int j=sum; j>=mon[i]; j--) {
                    dp[j] = max(dp[j], (1.0-pp[i])*dp[j-mon[i]]);
                }
            }
            for(i=sum; 1-dp[i]>=p; i--);
　　　　　　　prd(i);
        } else {//没有被抓的概率 
            prd(sum);
        }
    }
}

int main(){
    input();
    return 0;
}