Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P5030 长脖子鹿放置

P5030 长脖子鹿放置

按照套路,先二分图染色,自己手玩,不难发现,一个格点与其所能到达的另一个格点的横(或纵)坐标的奇偶性相反,那么我们可以按行(或列)的奇偶性将这些格点分成两个集合。

查看代码
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#include <iostream>
#include <cstdio>
#include <queue>
#include <climits>
#define INF INT_MAX
using namespace std;
const int N = 4e4 + 10, M = 2e6 + 10;
int dx[8] = { 1, 1, -1, -1, 3, 3, -3, -3 },
dy[8] = { 3, -3, 3, -3, 1, -1, 1, -1 };
bool vis[N];
int n, m, k, st, ed, tot, d[N], cur[N];
int idx = -1, hd[N], nxt[M], edg[M], wt[M];
bool bfs()
{
for (int i = st; i <= ed; i++)
d[i] = -1;
d[st] = 0;
cur[st] = hd[st];
queue <int> q;
q.push(st);
while (!q.empty())
{
int t = q.front();
q.pop();
for (int i = hd[t]; ~i; i = nxt[i])
if (d[edg[i]] == -1 && wt[i])
{
cur[edg[i]] = hd[edg[i]];
d[edg[i]] = d[t] + 1;
if (edg[i] == ed)
return true;
q.push(edg[i]);
}
}
return false;
}
int find(int x, int limit)
{
if (x == ed)
return limit;
int res = 0;
for (int i = cur[x]; ~i && res < limit; i = nxt[i])
{
cur[x] = i;
if (d[edg[i]] == d[x] + 1 && wt[i])
{
int t = find(edg[i], min(wt[i], limit - res));
if (!t)
d[edg[i]] = -1;
wt[i] -= t;
wt[i ^ 1] += t;
res += t;
}
}
return res;
}
int dinic()
{
int res = 0, flow;
while (bfs())
while (flow = find(st, INF))
res += flow;
return res;
}
void add(int x, int y, int z)
{
nxt[++idx] = hd[x];
hd[x] = idx;
edg[idx] = y;
wt[idx] = z;
}
bool inside(int x, int y)
{
return x > 0 && y > 0 && x <= n && y <= m;
}
int num(int x, int y)
{
return (x - 1) * m + y;
}
int main()
{
char a;
cin >> n >> m >> k;
tot = n * m;
st = 0;
ed = tot + 1;
for (int i = st; i <= ed; i++)
hd[i] = -1;
for (int i = 1, a, b; i <= k; i++)
{
cin >> a >> b;
int t = num(a, b);
if (vis[t])
continue;
vis[t] = true;
tot--;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
{
int t = num(i, j);
if (vis[t])
continue;
if (i & 1)
{
add(st, t, 1);
add(t, st, 0);
for (int k = 0; k < 8; k++)
{
int nx = i + dx[k],
ny = j + dy[k];
int h = num(nx, ny);
if (!inside(nx, ny) || vis[h])
continue;
add(t, h, INF);
add(h, t, 0);
}
}
else
{
add(t, ed, 1);
add(ed, t, 0);
}
}
cout << tot - dinic();
return 0;
}