【LOJ】#2115. 「HNOI2015」落忆枫音

题解

如果不加这条边，那么答案是所有点入度的乘积

加上了这条边之后，我们转而统计不合法的方案数

就是相当于统计一条路径从y到x，新图所有点度的乘积除上这条路径所有点的点度乘积

初始化为\(f[y] = \frac{\prod_{i = 2}^{n} ind[i]}{ind[y]}\)
转移按照拓扑序转移
如果u能到v
\(f[v] += \frac{f[u]}{ind[v]}\)

用总答案减掉f[x]即可

特判掉y = 1的情况

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef long double db;
typedef unsigned int u32;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
struct node {
    int to,next;
}E[MAXN * 2];
const int MOD = 1000000007;
int N,M,x,y,head[MAXN],sumE,ind[MAXN],f[MAXN],c[MAXN],inv[MAXN];
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
	if(c & 1) res = mul(res,t);
	t = mul(t,t);
	c >>= 1;
    }
    return res;
}
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
queue<int> Q;
void BFS() {
    Q.push(1);
    while(!Q.empty()) {
	int u = Q.front();Q.pop();
	for(int i = head[u] ; i ; i = E[i].next) {
	    int v = E[i].to;
	    f[v] = inc(f[v],mul(f[u],inv[v]));
	    --c[v];
	    if(!c[v]) Q.push(v);
	}
    }
}
void Solve() {
    read(N);read(M);read(x);read(y);
    int u,v;
    for(int i = 1 ; i <= M ; ++i) {
	read(u);read(v);
	add(u,v);ind[v]++;
    }
    if(y == 1) {
	int ans = 1;
	for(int i = 2 ; i <= N ; ++i) ans = mul(ans,ind[i]);
	out(ans);enter;
    }
    else {
	f[y] = 1;
	for(int i = 2 ; i <= N ; ++i) {
	    if(i != y) f[y] = mul(f[y],ind[i]);
	}
	for(int i = 1 ; i <= N ; ++i) {c[i] = ind[i];inv[i] = fpow(ind[i],MOD - 2);}
	BFS();
	int ans = 1;
	++ind[y];
	for(int i = 2 ; i <= N ; ++i) ans = mul(ans,ind[i]);
	ans = inc(ans,MOD - f[x]);
	out(ans);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}