Blog of RuSun

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

P3033 [USACO11NOV]Cow Steeplechase G

P3033 [USACO11NOV]Cow Steeplechase G

横的线段只与横的线段相交,竖的线段只与竖的线段相交,所以可以看做一个二分图。相交的线段只能选一个,最小割即可。

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 = 260, M = 2e5 + 10;
int n, st, ed, d[N], cur[N];
int idx = -1, hd[N], nxt[M], edg[M], wt[M];
struct Point
{
bool dir;
int x1, y1, x2, y2;
void input()
{
cin >> x1 >> y1 >> x2 >> y2;
if (x1 > x2)
swap(x1, x2);
if (y1 > y2)
swap(y1, y2);
if (x1 == x2)
dir = true;
else
dir = false;
}
}points[N];
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 exploit(int x, int limit)
{
if (x == ed)
return limit;
int res = 0;
for (int i = hd[x]; ~i && res <= limit; i = nxt[i])
{
cur[x] = i;
if (d[edg[i]] == d[x] + 1 && wt[i])
{
int t = exploit(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 = exploit(st, inf))
res += flow;
return res;
}
void add(int a, int b, int c)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
}
int main()
{
cin >> n;
st = 0;
ed = n + 1;
for (int i = st; i <= ed; i++)
hd[i] = -1;
for (int i = 1; i <= n; i++)
points[i].input();
for (int i = 1; i <= n; i++)
if (points[i].dir)
{
add(st, i, 1);
add(i, st, 0);
for (int j = 1; j <= n; j++)
if (!points[j].dir && points[i].x1 >= points[j].x1 && points[i].x1 <= points[j].x2 && points[j].y1 >= points[i].y1 && points[j].y1 <= points[i].y2)
{
add(i, j, inf);
add(j, i, 0);
}
}
for (int i = 1; i <= n; i++)
if (!points[i].dir)
{
add(i, ed, 1);
add(ed, i, 0);
}
cout << n - dinic();
return 0;
}