HDU 6064 RXD and numbers（BEST theorem）

RXD and numbers

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 40 Accepted Submission(s): 16

Problem Description
RXD has a sequence A1,A2,A3,…An, which possesses the following properties:
1≤Ai≤m
A1=An=1
for all 1≤x≤m, there is at least one position p where Ap=x.
for all x,y, the number of i(1 ≤ i < n) which satisfies Ai=x and Ai+1=y is Dx,y.
One day, naughty boy DXR clear the sequence.
RXD wants to know, how many valid sequences are there.
Output the answer module 998244353.
0≤Di,j<500,1≤m≤400
n≥2

Input
There are several test cases, please keep reading until EOF.
There are about 10 test cases, but only 1 of them satisfies m>50
For each test case, the first line consists of 1 integer m, which means the range of the numbers in sequence.
For the next m lines, in the i-th line, it consists of m integers, the j-th integer means Di,j.
We can easily conclude that n=1+∑mi=1∑mj=1Di,j.

Output
For each test case, output “Case #x: y”, which means the test case number and the answer.

Sample Input
2
1 2
2 1
4
1 0 0 2
0 3 0 1
2 1 0 0
0 0 3 1
4
0 1 0 0
1 0 0 0
0 0 0 1
0 0 1 0

Sample Output
Case #1: 6
Case #2: 18
Case #3: 0

Source
2017 Multi-University Training Contest - Team 3

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题目大意：

　给出一个m个节点的有向图中，每种起点终点边的条数，求有多少条从1号节点起始的欧拉回路。

解题思路：

　由于是要求有向图的欧拉回路数，很自然想到BEST theorem解决。
　BEST theorem的介绍引用wiki:

　这里要用到matrix tree的有向图版本，表达能力有限(:з」∠)，同样引用wiki：

　首先利用BEST theorm求得的欧拉回路数是不定起点的，这里固定起点为1，那么就需要把方案数乘上deg(1)，表示同一条欧拉回路，在这里起点不同算作不同的欧拉回路。由于BEST theorm会把重边看作不同的边，而本题会看作相同的边，所以还需要对答案除以∏mi=1∏mj=1(Di,j!)
　所以最终答案就是tw(G)∗(deg(1)!)∗∏mi=2(deg(i)−1)!/∏mi=1∏mj=1(Di,j)!
　总复杂度为O(m3)。
　

AC代码

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <ctime>#include <vector>#include <queue>#include <stack>#include <deque>#include <string>#include <map>#include <set>#include <list>using namespace std;#define INF 0x3f3f3f3f#define LL long long#define fi first#define se second#define mem(a,b) memset((a),(b),sizeof(a))const LL MOD=998244353;const int MAXV=400+1;int V;LL D[MAXV][MAXV];//从i，到j的边的数目LL in[MAXV],out[MAXV];//每个结点的入度，出度struct Matrix{    LL a[MAXV][MAXV];    Matrix()    {        memset(a,0,sizeof(a));    }    LL det(int n)//求前n行n列的行列式的值    {        for(int i=0;i<n;++i)            for(int j=0;j<n;++j)                a[i][j]=(a[i][j]%MOD+MOD)%MOD;        LL ret=1;        for(int i=0;i<n;i++)        {            for(int j=i+1;j<n;j++)                while(a[j][i])                {                    LL t=a[i][i]/a[j][i];                    for(int k=i;k<n;++k)                        a[i][k]=((a[i][k]-a[j][k]*t)%MOD+MOD)%MOD;                    for(int k=i;k<n;++k)                        swap(a[i][k],a[j][k]);                    ret=-ret;                }            if(!a[i][i])                return 0;            ret=ret*a[i][i]%MOD;        }        ret=(ret%MOD+MOD)%MOD;        return ret;    }};LL get_fac(LL x)//计算阶乘{    LL res=1;    for(LL i=2;i<=x;++i)        res=(res*i)%MOD;    return res;}LL exgcd(LL a, LL b, LL &x, LL &y){    LL d=a;    if(b)    {        d=exgcd(b, a%b, y, x);        y-=(a/b)*x;    }    else    {        x=1;        y=0;    }    return d;}LL inv(LL a)//计算逆元{    LL x, y;    exgcd(a, MOD, x, y);    return (MOD+x%MOD)%MOD;}void init()//初始化{    for(int i=0;i<=V;++i)        in[i]=out[i]=0;}int main(){    int cas=1;    while(~scanf("%d",&V))    {        init();        Matrix mat;        for(int i=0;i<V;++i)            for(int j=0;j<V;++j)            {                scanf("%lld", &D[i][j]);                mat.a[i][j]-=D[i][j];                mat.a[j][j]+=D[i][j];                in[j]+=D[i][j];                out[i]+=D[i][j];            }        //如果存在点入度不等于出度，则不存在欧拉回路直接输出0        bool ok=true;        for(int i=0;i<V;++i)            if(in[i]!=out[i])            {                ok=false;                break;            }        if(!ok)        {            printf("Case #%d: 0\n", cas++);            continue;        }        //把根节点移到最后，方便去掉它求行列式        for(int i=0;i<V;++i)            swap(mat.a[0][i], mat.a[V-1][i]);        for(int i=0;i<V;++i)            swap(mat.a[i][0], mat.a[i][V-1]);        LL ans=mat.det(V-1);        for(int i=0;i<V;++i)            ans=(ans*get_fac(in[i]-(i!=0)))%MOD;        for(int i=0;i<V;++i)            for(int j=0;j<V;++j)                ans=(ans*inv(get_fac(D[i][j])))%MOD;        printf("Case #%d: %lld\n", cas++, ans);    }    return 0;}