# bzoj 1412 [ZJOI2009]狼和羊的故事 最小割建图

### 解法

$S$集看作和羊连接，$T$看作和狼连接

### 代码

#include <bits/stdc++.h>
#define inf 1 << 30
#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 dx[5] = {0, -1, 1, 0, 0}, dy[5] = {0, 0, 0, -1, 1};
int n, m, s, t, cnt, a[N][N], l[N * N], cur[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) {
}
int calc(int x, int y) {
return (x - 1) * m + y;
}
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, inf);
}
return ret;
}
int main() {
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
s = 0, t = cnt = n * m + 1;
if (cnt % 2 == 0) cnt++;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++) {
int x = calc(i, j);
if (a[i][j] == 1) Add(s, x, inf);
if (a[i][j] == 2) {Add(x, t, inf); continue;}
for (int k = 1; k <= 4; k++) {
int tx = i + dx[k], ty = j + dy[k];
if (!tx || !ty || tx > n || ty > m) continue;
int y = calc(tx, ty);
if (a[tx][ty] == 2) Add(x, y, 1);
else if (!a[tx][ty]) Add(x, y, 1);
}
}
cout << dinic() << "\n";
return 0;
}


posted @ 2018-08-14 19:03  谜のNOIP  阅读(82)  评论(0编辑  收藏  举报