zoukankan      html  css  js  c++  java
  • Robberies(HDU 2955 DP01背包)

    Robberies

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 18013    Accepted Submission(s): 6653


    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
    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
    用逃脱概率比作价值,所抢钱数比作背包容量。
     1 #include <iostream>
     2 #include <cstring>
     3 #include <algorithm>
     4 using namespace std;
     5 float dp[10000+5];
     6 struct node
     7 {
     8     int a;
     9     float b;
    10 }f[105];
    11 int main()
    12 {
    13     int T,i,j,n;
    14     float p;
    15     freopen("in.txt","r",stdin);
    16     scanf("%d",&T);
    17     while(T--)
    18     {
    19         int sum=0;
    20         memset(dp,0,sizeof(dp));
    21         scanf("%f%d",&p,&n);
    22         for(i=1;i<=n;i++)
    23         {
    24             scanf("%d%f",&f[i].a,&f[i].b);
    25             f[i].b=1-f[i].b;
    26             sum+=f[i].a;
    27         }
    28         dp[0]=1.0;
    29         for(i=1;i<=n;i++)
    30         {
    31             for(j=sum;j>=f[i].a;j--)
    32             {
    33                 if(dp[j]<dp[j-f[i].a]*f[i].b)
    34                     dp[j]=dp[j-f[i].a]*f[i].b;
    35             }
    36         }
    37         for(i=sum;i>=0;i--)
    38             if(dp[i]>=(1.0-p))
    39                 break;
    40         printf("%d
    ",i);
    41     }
    42     return 0;
    43 }
  • 相关阅读:
    UVA
    Codeforces 898F 字符串hash
    牛客练习赛11 B trie树+拓扑判环 E 分治求平面最近点对
    Codeforces Round #459 (Div. 2) C 思维,贪心 D 记忆化dp
    2017-2018 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2017) D bfs思维 E bfs I Floyd找最小环 K 二分
    Educational Codeforces Round 37 E 补图求连通块 bfs+链表优化 F 线段树套路
    Codeforces 919E 数论,思维
    大数加法(主要是想记住模板)
    基础图论3
    简单并查集2hdu1213
  • 原文地址:https://www.cnblogs.com/a1225234/p/5173351.html
Copyright © 2011-2022 走看看