BZOJ 1412 [ZJOI2009]狼和羊的故事 | 网络流

时间:2023-03-08 16:31:25

显然是个最小割嘛!

一开始我是这么建图的:

  1. 源点向狼连INF
  2. 羊向汇点连INF
  3. 每两个相邻格子间连双向边,边权为1

然后T成狗

后来我是这么建图的:

  1. 源点向狼连INF
  2. 羊向汇点连INF
  3. 狼和空地向相邻的非狼节点连1

然后跑得 跟HK Journalist 跟中午抢饭的国大班同学们一样快

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
#define enter putchar('\n')
#define space putchar(' ')
template <class T>
void read(T &x){
char c;
bool op = 0;
while(c = getchar(), c > '9' || c < '0')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
} const int W = 105, N = 100005, M = 1000005, INF = 0x3f3f3f3f;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
int n, m, mp[W][W], src, des;
int ecnt = 1, adj[N], cur[N], dis[N], go[M], nxt[M], cap[M];
#define id(x, y) (((x) - 1) * m + (y)) void ADD(int u, int v, int _cap){
go[++ecnt] = v;
nxt[ecnt] = adj[u];
adj[u] = ecnt;
cap[ecnt] = _cap;
}
void add(int u, int v, int _cap){
ADD(u, v, _cap);
ADD(v, u, 0);
}
bool bfs(){
static int que[N], qr;
for(int i = 1; i <= des; i++)
dis[i] = -1, cur[i] = adj[i];
dis[src] = 0, que[qr = 1] = src;
for(int ql = 1; ql <= qr; ql++){
int u = que[ql];
for(int e = adj[u], v; e; e = nxt[e])
if(cap[e] && dis[v = go[e]] == -1){
dis[v] = dis[u] + 1, que[++qr] = v;
if(v == des) return 1;
}
}
return 0;
}
int dfs(int u, int flow){
if(u == des) return flow;
int ret = 0, delta;
for(int &e = cur[u], v; e; e = nxt[e])
if(cap[e] && dis[v = go[e]] == dis[u] + 1){
delta = dfs(v, min(flow - ret, cap[e]));
if(delta){
cap[e] -= delta;
cap[e ^ 1] += delta;
ret += delta;
if(ret == delta) return ret;
}
}
dis[u] = -1;
return ret;
}
int maxflow(){
int ret = 0;
while(bfs()){
int flow;
do{
ret += (flow = dfs(src, INF));
}while(flow);
}
return ret;
}
bool legal(int i, int j){
return i > 0 && j > 0 && i <= n && j <= m && mp[i][j] != 1;
} int main(){ read(n), read(m), src = n * m + 1, des = src + 1;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
read(mp[i][j]);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++){
if(mp[i][j] == 1) add(src, id(i, j), INF);
else if(mp[i][j] == 2) add(id(i, j), des, INF);
if(mp[i][j] != 2)
for(int k = 0, x, y; k < 4; k++)
if(legal(x = i + dx[k], y = j + dy[k]))
add(id(i, j), id(x, y), 1);
}
write(maxflow()), enter; return 0;
}