Blog of RuSun

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

P2526 [SHOI2001]小狗散步

P2526 [SHOI2001]小狗散步

题目有一点没有说清楚:同一个景点多次访问只算一次,或者说只能一个景点访问一次。

Pandog 每次与主人相遇之前最多只去一个景点。

即主人的一段路程中,小狗要以两倍速从起点到达一个景点再到终点,在主人到达之前完成。所以对于每段路程,和可到达的景点匹配即可。

查看代码
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#include <iostream>
#include <cstdio>
#include <queue>
#include <climits>
#define inf INT_MAX
using namespace std;
typedef pair <int, int> PII;
typedef long long LL;
const int N = 210, M = 3e4 + 10;
PII dog[N], site[N];
int n, m, st, ed, 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 exploit(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 = 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;
}
bool check (int x, int z1, int z2, int y)
{
LL a = z1 - site[x].first,
b = z2 - site[x].second,
c = z1 - site[y].first,
d = z2 - site[y].second;
LL t = 4 * ((a * a + b * b) * (c * c + d * d)),
h = 3 * (a * a + b * b + c * c + d * d ) - 8 * (a * c + b * d);
return t <= h * h;
}
int main ()
{
cin >> n >> m;
st = 0;
ed = n + m;
for (int i = st; i <= ed; i++)
hd[i] = -1;
for (int i = 1; i < n; i++)
{
add(st, i, 1);
add(i, st, 0);
}
for (int i = 1; i <= n; i++)
cin >> site[i].first >> site[i].second;
for (int i = 1, a, b; i <= m; i++)
{
cin >> dog[i].first >> dog[i].second;
for (int j = 1; j < n; j++)
if (check(j, dog[i].first, dog[i].second, j + 1))
{
add(j, i + n - 1, 1);
add(i + n - 1, j, 0);
}
}
for (int i = n; i < ed; i++)
{
add(i, ed, 1);
add(ed, i, 0);
}
cout << n + dinic() << endl;
for (int i = 1; i < n; i++)
{
cout << site[i].first << ' ' << site[i].second << ' ';
for (int j = hd[i]; ~j; j = nxt[j])
if (edg[j] != st && !wt[j])
cout << dog[edg[j] - n + 1].first << ' ' << dog[edg[j] - n + 1].second << ' ';
}
cout << site[n].first << ' ' << site[n].second;
return 0;
}