ZOJ3319 DP 通过入度出度判非法情况

将自己历史的AC共享

zoj3319 DP题

通过入度出度判非法情况，dp[i]+=c[i-1][j]*only[j+1]*dp[i-1-j] (1<=j<=i-3) dp[i]+=only[i]; 以上是单独结点的情况，然后将图转化，统计连通分量数(n)，和个数不为1的连通分量数(edge),然后ans+=c[edge][i]*dp[n-i] (1<=i<=edge) ans+=dp
;

//2157221 2010-04-12 20:23:46 Accepted  3319 C++ 10 240 greentea@FireStar
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <ctime>
#include <utility>
#include <stdexcept>

using namespace std;

#define inf (1<<30)
#define PB push_back
#define mset(x,a) memset(x,(a),sizeof(x))
#define SIZE(X) ((int)X.size())
#define dbg(x) cerr<<#x<<" : "<<x<<endl
#define rg(x,y) (x=x>y?x:y)
typedef vector<int> VI;
typedef vector<char> VC;
typedef vector<string> VS;
typedef long long LL;
typedef unsigned long long uLL;
#define twoL(X) (((LL)(1))<<(X))
const double PI=acos(-1.0);
const double eps=1e-11;
template <class T> T sqr(T x) {return x*x;}
template <class T> T gcd(T a, T b) {if(a<0) return (gcd(-a,b)); if(b<0) return (gcd(a,-b)); return (b==0)?a:gcd(b,a%b);}
template <class T> T lcm(T a, T b) {return a*b/gcd(a,b);}
LL toLL(string s) { istringstream sin(s); LL t; sin>>t; return t;}
int toInt(string s) {istringstream sin(s); int t; sin>>t; return t;}
string toString(LL v) {ostringstream sout; sout<<v; return sout.str();}
string toString(int v) {ostringstream sout; sout<<v; return sout.str();}
#define FOREACH(it, a) for(typeof((a).begin()) it = (a).begin(); it!=(a).end(); ++it)
#define ALL(x) ((x).begin(), (x).end())
#define cross(a, b, c)  ((c).x-(a).x)*((b).y-(a).y)-((b).x-(a).x)*((c).y-(a).y)
#define sq_dist(p, q) ((p).x-(q).x)*((p).x-(q).x)+((p).y-(q).y)*((p).y-(q).y)
#define FF(i,n) for(int i = 0; i < n; ++i)

#define mod 10000007
int r, n;
char mat[105][105];
int dp[105];
int only[105];
int c[105][105];
int M[105];
int in[105], out[105];
bool v[105], e[105];

int modMul(int a, int b, int m) {
int t=0;
a=(a%m+m)%m;
b=(b%m+m)%m;
while(b){
if(b&1){
t=(t+a)%m;
}
a=(a+a)%m;
b>>=1;
}
return t;
}

void init() {
mset(c, 0);
c[0][0] = 1;
for( int i = 1; i <= 100; ++i )
c[i][0] = 1;
for( int i = 1; i <= 100; ++i ) {
for( int j = 1; j <= i; ++j )
c[i][j] = (c[i-1][j-1]+c[i-1][j])%mod;
}
}

void workDp() {
mset(dp, 0);
dp[0] = 1, dp[1] = 0, dp[2] = 1, dp[3] = 2;
only[2] = 1;
only[3] = 2;
for( int i = 4; i <= 100; ++i ) {
only[i] = only[i-1]*(i-1)%mod;
dp[i] = only[i];
for( int j = 1; j <= (i-3); ++j ) {
int tt;
tt = modMul(c[i-1][j],only[j+1], mod);
tt = modMul(tt, dp[i-1-j], mod);
dp[i] += tt;
dp[i] %= mod;
}
}
}

int main()
{
init();
workDp();
while( scanf("%d", &r), r ) {
mset(in, 0);
mset(out, 0);
mset(v, 0);
mset(e, 0);
for( int i = 0; i < r; ++i ) {
M[i] = i;
scanf("%s", mat[i]);
for( int j = 0; j < r; ++j ) {
if( mat[i][j] == 'Y' )
in[j]++, out[i]++;
}
}
int flag = 0;
for( int i = 0; i < r; ++i ) {
if( in[i] > 1 || out[i] > 1 )   flag = 1;
}
if( flag ) {
printf("0\n");
continue;
}
n = 0;
for( int i = 0; i < r; ++i ) {
for( int j = 0; j < r; ++j ) {
if( mat[i][j] == 'Y' )    M[i] = j;
}
}
int edge = 0;
for( int i = 0; i < r; ++i ) {
int k  = i, t = i, flag = 0;
while( M[k] != k ) {
k = M[k];
if( k == t ) {
v[k] = 1;
flag = 1;
break;
}
}
if( M[i] != i && e[k] == 0 && flag == 0 ) {
e[k] = 1;
edge++;
}
if( v[k] )    continue;
v[k] = 1;
n++;
}
int ans = 0;
for( int i = 1; i <= edge; ++i ) {
ans += modMul(c[edge][i], dp[n-i], mod);
ans %= mod;
}

ans = (ans+dp
)%mod;
printf("%d\n", ans);
}
return 0;
}

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