Regionals >> Asia-tehran >> 7527-funfair "greedy" "DP"

7527-funfair

Topic Links: 7527

The main topic: play a Level game, the initial X-yuan, a total of n off, their choice of K-off, as well as the order of clearance. When I close, win the probability of Pi, won the Ai Yuan after winning, lost, then lose Li * x money. Ask how to choose off and the order of the level so that the final amount of money to expect the most.

Topic idea: First, need to sort, this can ensure that the first i+1 off after I close, and then to DP, I close or not to take.

Sort by: We can know that when I close the first

赢: (Ai + x) * Pi输: (1 - Pi)(1 - Li) * x相加合并一下得到（1 - Li + LiPi） * x + Ai * Pi假设 c = （1 - Li + LiPi）    d = Ai * Pi即: cx + d那么，在考虑先过A关还是B关的时候，有两种可能性，先过A关：c2 * (c1*x+d1) + d2;先过B关：c1 * (c2*x+d2) + d1;假设A大于B，c2 * (c1*x+d1) + d2 > c1 * (c2*x+d2) + d1化简得到，d1/(c1-1) < d2/(c2-1)

So: Sort by d1/(c1-1) < d2/(c2-1).

Attention:
1. When sorting, you need to consider the case of C equals 1 (denominator cannot be 0)
When the 2.p,l is in, divide by 100 (because they are percentages)

The last DP to handle it!

Here's the code:

#include <iostream>#include <iomanip>#include <fstream>#include <sstream>#include <cmath>#include <cstdio>#include <cstring>#include <cctype>#include <algorithm>#include <functional>#include <numeric>#include <string>#include <set>#include <map>#include <stack>#include <vector>#include <queue>#include <deque>#include <list>using namespace STD;#define EPS 1e-6structnode{DoubleA,l,p,c,d,s;} g[ the];BOOLCMP (Node A, Node B) {if(A.C = =1|| B.C = =1)returnA.C < B.C;//Compare pits!     returnA.S > B.S;}Doubledp[ the][ the];intMain () {intN,k,x; while(Cin>> N >> k >> x) {if(n = =0&& k = =0&& x = =0) Break;memset(DP,0,sizeof(DP));memset(g,0,sizeof(g)); for(inti =0; I < n; i++) {Cin>> g[i].a >> g[i].l >> g[i].p; G[I].L/= -; G[I].P/= -; G[I].C =1-G[I].L + g[i].l * G[I].P;            G[I].D = g[i].a * G[I].P; G[I].S = G[I].D/(G[I].C-1); } sort (g,g + n,cmp); for(inti =0; I < n; i++) dp[i][0] = x; for(inti =0; I < n; i++)//I choose not to select an activity{dp[i][1] = max (x * g[i].c + g[i].d,dp[i][1]); for(intL =0; l < i; l++)//In the first L activities{ for(intj =2; J <= min (l +2, k); J + +)//Selected J{Dp[i][j] = max (Dp[i][j],dp[l][j-1] * g[i].c + G[I].D); }            }        }DoubleAns =0; for(inti =0; I < n; i++) ans = max (Dp[i][k],ans);printf("%.2f\n", ans); }}

Attach several sets of test data:

/ * input * /5 3  - $  the  - the  $ 7 A  in Panax Notoginseng -  the  $ -  About  -Ten 5  + -  the  - in  A  A $  -  - the 98  the -  -  at $ 98  the -  -  the the  the  the -  the  the $  A  + - 1 Ten A 94  +Ten  at  the -  -  the -  -  the -  -  Wu the  -  the A  -  - -  the  the $ 4  * -  the  the *  the  + the  the  - +  -  - -  -  - the 4  thePanax Notoginseng  the  - -  -  - the  - Ten in  the  A - 7  the - 5 Ten in  -  A Wu 8  - -  in  - the  the 3 -  A  in the 4  - the 2  the the  the  the the  A  - -  -  - the  -  - -  -  the the  One  the -  - 94 -  -  the the  the 3 -  -  + at  the  + $  the  About - 0  the/ * Output * /90.67860.3979.80213.04

Regionals >> Asia-tehran >> 7527-funfair "greedy" "DP"