Description
Everyone knows of the secret agent double-oh-seven, the popular Bond (James Bond). A lesser known fact is that he actually did not perform most of his missions by himself; they were instead done by his cousins, Jimmy Bonds. Bond (James Bond) has grown weary of having to distribute assign missions to Jimmy Bonds every time he gets new missions so he has asked you to help him out. Every month Bond (James Bond) receives a list of missions. Using his detailed intelligence from past missions, for every mission and for every Jimmy Bond he calculates the probability of that particular mission being successfully completed by that particular Jimmy Bond. Your program should process that data and find the arrangement that will result in the greatest probability that all missions are completed successfully. Note: the probability of all missions being completed successfully is equal to the product of the probabilities of the single missions being completed successfully.
有(n)个人去执行(n)个任务，每个人执行每个任务有不同的成功率，每个人只能执行一个任务，求所有任务都执行的总的成功率。
输入第一行，一个整数(n)（(1leq nleq 20) ），表示人数兼任务数。接下来(n)行每行(n)个数，第(i)行第(j)个数表示第(i)个人去执行第(j)个任务的成功率（这是一个百分数，在(0)到(100)间）。
输出最大的总成功率（这应也是一个百分数）
Input
The first line will contain an integer N, the number of Jimmy Bonds and missions (1 ≤ N ≤ 20). The following N lines will contain N integers between 0 and 100, inclusive. The j-th integer on the ith line is the probability that Jimmy Bond i would successfully complete mission j, given as a percentage.
Output
Output the maximum probability of Jimmy Bonds successfully completing all the missions, as a percentage.
Sample Input 1
2
100 100
50 50
Sample Output 1
50.000000
Sample Input 2
2
0 50
50 0
Sample Output 2
25.00000
Sample Input 3
3
25 60 100
13 0 50
12 70 90
Sample Output 3
9.10000
HNIT
Clarification of the third example: If Jimmy bond 1 is assigned the 3rd mission, Jimmy Bond 2 the 1st mission and Jimmy Bond 3 the 2nd mission the probability is: 1.0 0.13 0.7 = 0.091 = 9.1%. All other arrangements give a smaller probability of success. Note: Outputs within ±0.000001 of the official solution will be accepted.
一看就是状压……（KM也能写，不过不想填坑了）
设(f[i][sta])表示前(i)个人所做任务状态为(sta)的成功率，转移就随便枚举一下即可
/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
#define lowbit(x) ((x)&-(x))
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
static char buf[1000000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
int x=0,f=1; char ch=gc();
for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<3)+(x<<1)+ch-'0';
return x*f;
}
inline int read(){
int x=0,f=1; char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<3)+(x<<1)+ch-'0';
return x*f;
}
inline void print(int x){
if (x<0) putchar('-'),x=-x;
if (x>9) print(x/10);
putchar(x%10+'0');
}
int g[(1<<20)+10],V[25][25];
double f[(1<<20)+10];
int main(){
int n=read();
for (int i=1;i<1<<n;i++) g[i]=g[i-lowbit(i)]+1;
for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) V[i][j]=read();
f[0]=1.0;
for (int i=1;i<=n;i++){
for (int sta=0;sta<1<<n;sta++){
if (g[sta]!=i) continue;
for (int j=1;j<=n;j++)
if (sta&(1<<(j-1)))
f[sta]=max(f[sta],f[sta^(1<<(j-1))]*V[i][j]/100);
}
}
printf("%lf
",f[(1<<n)-1]*100);
return 0;
}