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【BZOJ1001】[BeiJing2006]狼抓兔子

2017-03-18 10:45 225 查看
挺简单一个题,最小割模板

我的感觉就是可能建图的时候会比较麻烦吧,毕竟三个方向。

#include <cctype>
#include <climits>
#include <cstdio>
#include <cstring>
#include <iostream>

#define debug(x) std::cout << #x << " = " << x << std::endl;

#define __OPTIMIZED__INPUT__
int nextInt()
{
int num = 0;
char c;
bool flag = false;
while ((c = std::getchar()) == ' ' || c == '\r' || c == '\n' || c == '\t');
if (c == '-')
flag = true;
else
num = c - 48;
while (std::isdigit(c = std::getchar()))
num = num * 10 + c - 48;
return (flag ? - 1 : 1) * num;
}

struct
{
int to;
int nex;
int v;
} e[600001];
int head[600001];
int h[600001], q[600001];
int ans, n, m;

void Insert(const int u, const int v, const int w)
{
static int tot = 0;
tot++;
e[tot].to = v;
e[tot].v = w;
e[tot].nex = head[u];
head[u] = tot;
}

bool bfs()
{
int now, i;
std::memset(h, 0xff, sizeof h);
int t = 0;
int w = 1;
q[t] = 1;
h[1] = 0;
while (t < w)
{
now = q[t];
t++;
i = head[now];
for (int i = head[now]; i; i = e[i].nex)
if (e[i].v && h[e[i].to] < 0)
{
q[w++] = e[i].to;
h[e[i].to] = h[now] + 1;
}
}
if (h[n * m] == -1)
return 0;
return 1;
}

int dfs(const int x, const int f)
{
if (x == n * m)
return f;
int w, used = 0;
for (int i = head[x]; i; i = e[i].nex)
if (e[i].v && h[e[i].to] == h[x] + 1)
{
w = dfs(e[i].to, std::min(w, e[i].v));
e[i].v -= w;
e[i + 1].v += w;
used += w;
if (used == f)
return f;
}
if (!used)
h[x] = -1;
//  debug(used);
return used;
}

void Dinic()
{
while (bfs())
ans += dfs(1, INT_MAX);
}

int main(int argc, char ** argv)
{
n = nextInt();
m = nextInt();
int x;
for (int i = 1; i <= n; i++)
for (int j = 1; j < m; j++)
{
x = nextInt();
Insert(m * (i - 1) + j, m * (i - 1) + j + 1, x);
Insert(m * (i - 1) + j + 1, m * (i - 1) + j, x);
}
for (int i = 1; i < n; i++)
for (int j = 1; j <= m; j++)
{
x = nextInt();
Insert(m * (i - 1) + j, m * i + j, x);
Insert(m * i + j, m * (i - 1) + j, x);
}
for (int i = 1; i < n; i++)
for (int j = 1; j < m; j++)
{
x = nextInt();
Insert(m * (i - 1) + j, m * i + j + 1, x);
Insert(m * i + j + 1, m * (i - 1) + j, x);
}
Dinic();
std::cout << ans/* << std::endl*/;
#ifdef __NOTEPADPP
std::cin.get();
std::cin.get();
#endif
return 0;
}


4000
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