HDU2955概率背包

Robberies

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

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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

30.04 31 0.022 0.033 0.050.06 32 0.032 0.033 0.050.10 31 0.032 0.023 0.05

Sample Output

246

Source

IDI Open 2009 

#define DeBUG#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <string>#include <set>#include <sstream>#include <map>#include <list>#include <bitset>using namespace std ;#define zero {0}#define INF 0x3f3f3f3f#define EPS 1e-6#define TRUE true#define FALSE falsetypedef long long LL;const double PI = acos(-1.0);//#pragma comment(linker, "/STACK:102400000,102400000")inline int sgn(double x){    return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);}#define N 100005double dp[N];//获取n钱最大不被抓概率struct node{    int money;    double p;};node nn[N];int main(){#ifdef DeBUGs    freopen("C:\\Users\\Sky\\Desktop\\1.in", "r", stdin);#endif    int T;    scanf("%d", &T);    while (T--)    {        double p;        int n;        scanf("%lf%d", &p, &n);        p = 1 - p;//不被抓概率        int v = 0;        for (int i = 0; i < n; i++)        {            scanf("%d%lf", &nn[i].money, &nn[i].p);            v += nn[i].money;            nn[i].p = 1 - nn[i].p;<span style="font-family: Arial, Helvetica, sans-serif;">//不被抓概率</span>        }        memset(dp, 0, sizeof(dp));        dp[0] = 1;        for (int i = 0; i < n; i++)        {            for (int j = v; j >= nn[i].money; j--)            {                dp[j] = max(dp[j], dp[j - nn[i].money] * nn[i].p);            }        }        // for (int i = 0; i < 10; i++)        // {        //     cout << dp[i] << endl;        // }        int maxx = 0;        for (int i = v; i >= 0; i--)        {            if (sgn(dp[i] - p) > 0)            {                maxx = i;                break;            }        }        printf("%d\n", maxx);    }    return 0;}