# bzoj 1143 [CTSC2008]祭祀river 网络流

### 解法

$x$与它所有能到达的点$y$连边，然后这依然是一个DAG

### 代码

#include <bits/stdc++.h>
#define N 110
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
x = 0; int f = 1; char c = getchar();
while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Edge {
int next, num, c;
} e[N * N * N];
int s, t, cnt, l[N * 2], cur[N * 2], a[N][N];
void add(int x, int y, int c) {
e[++cnt] = (Edge) {e[x].next, y, c};
e[x].next = cnt;
}
void Add(int x, int y, int c) {
}
bool bfs(int s) {
for (int i = 1; i <= t; i++) l[i] = -1;
queue <int> q; q.push(s);
while (!q.empty()) {
int x = q.front(); q.pop();
for (int p = e[x].next; p; p = e[p].next) {
int k = e[p].num, c = e[p].c;
if (c && l[k] == -1)
l[k] = l[x] + 1, q.push(k);
}
}
return l[t] != -1;
}
int dfs(int x, int lim) {
if (x == t) return lim;
int used = 0;
for (int p = cur[x]; p; p = e[p].next) {
int k = e[p].num, c = e[p].c;
if (c && l[k] == l[x] + 1) {
int w = dfs(k, min(c, lim - used));
e[p].c -= w, e[p ^ 1].c += w; used += w;
if (e[p].c) cur[x] = p;
if (used == lim) return lim;
}
}
if (!used) l[x] = -1; return used;
}
int dinic() {
int ret = 0;
while (bfs(s)) {
for (int i = 0; i <= t; i++) cur[i] = e[i].next;
ret += dfs(s, INT_MAX);
}
return ret;
}
int main() {
for (int i = 1; i <= m; i++) {
a[x][y] = 1;
}
for (int k = 1; k <= n; k++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (i != k && i != j && j != k)
a[i][j] |= a[i][k] & a[k][j];
s = 0, t = cnt = 2 * n + 1;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (a[i][j] && i != j) Add(i, j + n, 1);
for (int i = 1; i <= n; i++)