HDU 6064 RXD and numbers(BEST theorem)
2017-08-02 11:09
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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
Recommend
liuyiding
BEST theorem的介绍引用wiki:
![](http://img.blog.csdn.net/20170802103716020?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvWWFzb2xh/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast)
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:
![](http://img.blog.csdn.net/20170802103827079?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvWWFzb2xh/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/SouthEast)
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)。
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
Recommend
liuyiding
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; }
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