HDU 2955 推薦 搶銀行+01揹包變形

HDU 2955 推薦

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

SubmitStatusPracticeHDU 2955

Appoint description:

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

解題思路：
以前做過一個浮點數的01揹包，開始感覺和那一題類似的思路，把浮點數轉化成整數來做
實際上不是，這題的精度大了去了，抱着試一試的態度果然wa了一發
1，轉換思路，重新設計狀態轉移方程
2，設計的思路是：搶到錢i時，不被抓的概率dp[i]
3，dp[0] = 1;狀態轉移方程就是dp[j] = max(dp[j],dp[j-v[i]]*p[i]) ;   
4，代碼中P和p[]數組都是拿1減了一下，表示不被抓的概率
5，最後在循環結束後的dp數組中找到第一個大於等於P的輸出即可
仔細看看那個方程，很好理解吧

#include<bits/stdc++.h>
using namespace std;
const int maxn = 105 ;
int v[maxn];
double p[maxn];
double dp[10005];
int main(){
    int T,n;
    double P ;
    scanf("%d",&T);
    while(T--){
        scanf("%lf %d",&P,&n) ;
        P = 1-P ;
        int tot= 0 ;
        for(int i=1;i<=n;i++){
            scanf("%d%lf",&v[i],&p[i]);
            p[i] = 1-p[i] ;
            tot+=v[i] ;
        }
        memset(dp,0,sizeof(dp));
        dp[0] = 1 ;
        for(int i=1;i<=n;i++){
            for(int j=tot;j>=v[i];j--){
                dp[j] = max(dp[j],dp[j-v[i]]*p[i]) ;
            }
//            for(int j = 0;j<=tot;j++){
//                printf("%lf ",dp[j]);
//            }printf("\n");
        }
        int ans = 0 ;
        for(int i = tot;i>=0;i--){
            //printf("%lf %lf\n",dp[i],P);
            if(dp[i]>=P){
                ans = i ;
                break;
            }
        }
        printf("%d\n",ans) ;
    }
    return 0;

}