HDU 2955 Robberies 01背包

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=2955

Robberies

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others) #### 问题描述 > 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.

输入

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 .

输出

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.

样例输入

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

样例输出

2
4
6

题意

小偷偷第i家银行能获利mi,被抓的概率为pi，各银行被抓概率独立，如果小偷被抓概率超过p，小偷就被抓，问小偷不被抓的最大获利。

题解

首先，概率没法做容量，我们只能考虑用钱做容量，dp[i]表示获利为i的最小被抓概率。
注意：已知i银行被抓概率为pi，j银行被抓概率为pj，则同时偷两银行概率为：pi+pj-pi*pj。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long  LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-6;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=11111;

int n,mon[111];
double pr,pro[111];
double dp[maxn];

int main() {
    int tc;
    scf("%d",&tc);
    while(tc--){
        scf("%lf%d",&pr,&n);
        for(int i=1;i<=n;i++){
            scf("%d%lf",mon+i,pro+i);
        }
        rep(i,0,maxn) dp[i]=2.0;
        dp[0]=0.0;
        for(int i=1;i<=n;i++){
            for(int j=maxn-1;j>=mon[i];j--){
                dp[j]=min(dp[j],dp[j-mon[i]]+pro[i]-dp[j-mon[i]]*pro[i]);
            }
        }
        int ans=0;
        for(int j=maxn-1;j>=0;j--){
            if(dp[j]<=eps+pr){
                ans=j; break;
            }
        }
        prf("%d
",ans);
    }
    return 0;
}

//end-----------------------------------------------------------------------

Notes

这题应该注意到独立概率的关键字！！！我竟然没有容斥直接拿去加了，概率渣xrz.