Blog of RuSun

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

P2472 [SCOI2007]蜥蜴

P2472 [SCOI2007]蜥蜴

可以理解,石柱的高度即可以经过的蜥蜴的数量的最大值,考虑拆点,将石柱拆为两点,流量为石柱的高度。所有蜥蜴的位置连接源点,流量为 $1$ 。枚举任意两个石柱,如果满足距离限制,则连流量为 $inf$ 的边。能到达边界外的石柱,向汇点连接流量为 $inf$ 的边。

查看代码
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#include <iostream>
#include <cstdio>
#include <queue>
#include <map>
#include <climits>
#define INF INT_MAX
using namespace std;
const int N = 4100, M = 8e5 + 10;
int n, R, C, D, st, ed, tot, d[N], cur[N];
int idx = -1, hd[N], nxt[M], edg[M], wt[M];
char mp[N][N];
pair <int, int> site[N];
map <pair <int, int>, int> num;
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 check (pair <int, int> x, pair <int, int> y)
{
int dx = x.first - y.first,
dy = x.second - y.second;
return dx * dx + dy * dy <= D * D;
}
int main ()
{
char c;
cin >> R >> C >> D;
for (int i = 1; i <= R; i++)
for (int j = 1; j <= C; j++)
{
cin >> mp[i][j];
if (mp[i][j] == '0')
continue;
site[++n] = make_pair(i, j);
num[make_pair(i, j)] = n;
}
st = 0;
ed = n << 1 | 1;
for (int i = st; i <= ed; i++)
hd[i] = -1;
for (int i = 1; i <= n; i++)
{
int tx = site[i].first,
ty = site[i].second;
add(i, i + n, mp[tx][ty] - '0');
add(i + n, i, 0);
if (tx - D < 1 || ty - D < 1 || tx + D > R || ty + D > C)
{
add(i + n, ed, INF);
add(ed, i + n, 0);
}
}
for (int i = 1; i <= R; i++)
for (int j = 1; j <= C; j++)
{
cin >> c;
if (c != 'L')
continue;
tot++;
int t = num[make_pair(i, j)];
add(st, t, 1);
add(t, st, 0);
}
for (int i = 1; i <= n; i++)
for (int j = i + 1; j <= n; j++)
if (check(site[i], site[j]))
{
add(i + n, j, INF);
add(j, i + n, 0);
add(j + n, i, INF);
add(i, j + n, 0);
}
cout << tot - dinic();
return 0;
}