随机算法我也只是稍微接触了一下，就是想写篇博客自己稍微总结一下

其实随机算法也算是一个玄学吧，运气不好还是会wa。但是我们知道，计算机可以在短时间内计算大量的数据，所以碰到正确答案的概率还是挺大的。

当然随机不是随机答案啦，对于不同的题有可以不同的随机对象。通常我们会在一下情况下用随机：
1、逐个枚举太多无用的情况下选择随机 以减少枚举次数
2、判断正确性的时候降低暴力判断的复杂度

下面还是放题吧

poj2454 Jersey Politics

Description
In the newest census of Jersey Cows and Holstein Cows, Wisconsin cows have earned three stalls in the Barn of Representatives. The Jersey Cows currently control the state’s redistricting committee. They want to partition the state into three equally sized voting districts such that the Jersey Cows are guaranteed to win elections in at least two of the districts.
Wisconsin has 3*K (1 <= K <= 60) cities of 1,000 cows, numbered 1..3*K, each with a known number (range: 0..1,000) of Jersey Cows. Find a way to partition the state into three districts, each with K cities, such that the Jersey Cows have the majority percentage in at least two of districts.
All supplied input datasets are solvable.
Input
* Line 1: A single integer, K
* Lines 2..3*K+1: One integer per line, the number of cows in each city that are Jersey Cows. Line i+1 contains city i’s cow census.
Output
* Lines 1..K: K lines that are the city numbers in district one, one per line
* Lines K+1..2K: K lines that are the city numbers in district two, one per line
* Lines 2K+1..3K: K lines that are the city numbers in district three, one per line
Sample Input
2
510
500
500
670
400
310
Sample Output
1
2
3
6
5
4
Hint
Other solutions might be possible. Note that “2 3” would NOT be a district won by the Jerseys, as they would be exactly half of the cows.
translate
在泽西奶牛和荷斯坦奶牛的最新普查中，威斯康星奶牛在谷仓中获得了三个档位。 泽西奶牛目前控制着国家重新分配委员会。 他们想将国家分为三个相当大的投票地区，以便保证泽西奶牛在至少两个地区获得选举权。
威斯康星州有3 * K（1 <= K = = 60）个1000头牛的城市，编号为1..3 * K，每头都有泽西奶牛的已知数量（范围：1.001）。 找到一种将州划分为三个区域的方式，每个区域都有K个城市，使得泽西奶牛在至少两个地区拥有多数比例。
所有提供的输入数据集都可以解决。

首先有一个贪心：把序列从大到小排序，取前2*k个作为需要满足要求的两个子序列的元素。之后就是对于这2*k个元素如何分配的问题了
如果dfs的话会出事（这是肯定的），那么如何减少枚举次数呢？就是用随机算法了
但是如何随机也是有讲究的。如果我们直接随机这个序列，很有可能两个子序列的元素并没有改变（只是交换了一下顺序）。我们希望每一次的枚举都有效，即交换元素。那么很显然，随机这个交换的元素就好了

#include<ctime>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;struct node{int val,num;
}a[200];
int k;bool cmp(const node &x,const node &y){return x.val>y.val;
}
int main(){srand((unsigned)time(NULL));while(scanf("%d",&k)!=EOF){for(int i=0;i<3*k;i++){scanf("%d",&a[i].val);a[i].num=i;}sort(a+0,a+3*k,cmp);int sum1=0,sum2=0;for(int i=0;i<k;i++){sum1+=a[i].val;sum2+=a[i+k].val;}while(1){if(sum1>500*k&&sum2>500*k) break;int r1=rand()%k,r2=rand()%k;sum1=sum1-a[r1].val+a[k+r2].val;sum2=sum2-a[k+r2].val+a[r1].val;swap(a[r1],a[k+r2]);}for(int i=0;i<3*k;i++) printf("%d\n",a[i].num+1);}return 0;
}

poj3318 Matrix Multiplication

Description
You are given three n × n matrices A, B and C. Does the equation A × B = C hold true?
Input
The first line of input contains a positive integer n (n ≤ 500) followed by the the three matrices A, B and C respectively. Each matrix’s description is a block of n × n integers.
It guarantees that the elements of A and B are less than 100 in absolute value and elements of C are less than 10,000,000 in absolute value.
Output
Output “YES” if the equation holds true, otherwise “NO”.
Sample Input
2
1 0
2 3
5 1
0 8
5 1
10 26
Sample Output
YES
Hint
Multiple inputs will be tested. So O(n3) algorithm will get TLE.
translate
给定三个n×n矩阵A，B和C.方程A×B = C是否成立？

题目的hint说了，暴力的把矩阵乘起来要炸。
那么如果可以找到一个快速的判断方法，不一定要对错两面都具有绝对性，只要其中一面具有绝对性就好。
设v为随机生成的n维列向量，每个元素取1或0。看A*(B*v)和C*v是否相同。但是这样不稳定，因为即使A*B不等于C，这样算出来可能是相等的。经计算，这种情况还相等的概率小于1/2。那么多判断几次，出错的概率就会大大降低。所以试60次就差不多了

#include<ctime>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;int n;
struct matrix{int k[505][505];int x,y;
}A,B,C,v,_B,_C,_A;void get_rand(matrix &v){for(int i=1;i<=n;i++) v.k[i][1]=rand()%2;
}
void mult_mt(matrix &_a,const matrix &a,const matrix &v){_a.x=a.x,_a.y=v.y;for(int i=1;i<=n;i++){int tmp=0;for(int j=1;j<=n;j++)tmp+=a.k[i][j]*v.k[j][1];_a.k[i][1]=tmp;}
}
bool check(const matrix &a,const matrix &b){for(int i=1;i<=n;i++) if(a.k[i][1]!=b.k[i][1]) return false;return true;
}
int main(){srand((unsigned)time(NULL));scanf("%d",&n);for(int i=1;i<=n;i++)for(int j=1;j<=n;j++) scanf("%d",&A.k[i][j]);for(int i=1;i<=n;i++)for(int j=1;j<=n;j++) scanf("%d",&B.k[i][j]);for(int i=1;i<=n;i++)for(int j=1;j<=n;j++) scanf("%d",&C.k[i][j]);A.x=A.y=B.x=B.y=C.x=C.y=n;int m=60;   v.x=n,v.y=1;while(m--){get_rand(v);mult_mt(_B,B,v);mult_mt(_C,C,v);mult_mt(_A,A,_B);if(!check(_C,_A)){printf("NO\n");return 0;}}printf("YES\n");return 0;
}