Blog of RuSun

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

P4843 清理雪道

P4843 清理雪道

可以从任意点开始,任意点结束,所以从源点向每个点连接流量为 $inf$ 的边,每个点向汇点连接流量为 $inf$ 的边,每条题目中给定的边的流量下限为 $1$ ,做上下界最小流即可。

查看代码
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#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 110, M = 5e4 + 10, inf = 1e4;
int n, st, ed, d[N], cur[N], o[N];
int idx = -1, hd[N], nxt[M], edg[M], wt[M];
bool bfs()
{
for (int i = 0; i <= n + 3; 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])
{
d[edg[i]] = d[t] + 1;
cur[edg[i]] = hd[edg[i]];
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(limit - res, wt[i]));
if (!t)
d[edg[i]] = -1;
res += t;
wt[i] -= t;
wt[i ^ 1] += 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()
{
scanf("%d", &n);
int s = n + 1, t = n + 2;
st = 0, ed = n + 3;
for (int i = 0; i <= n + 3; i++)
hd[i] = -1;
for (int i = 1, m; i <= n; i++)
{
scanf("%d", &m);
add(s, i, inf);
add(i, s, 0);
add(i, t, inf);
add(t, i, 0);
for (int a; m; m--)
{
scanf("%d", &a);
add(i, a, inf);
add(a, i, 0);
o[i]--;
o[a]++;
}
}
for (int i = 1; i <= n; i++)
if (o[i] < 0)
{
add(i, ed, -o[i]);
add(ed, i, 0);
}
else if (o[i] > 0)
{
add(st, i, o[i]);
add(i, st, 0);
}
add(t, s, inf);
add(s, t, 0);
dinic();
int res = wt[idx];
wt[idx] = wt[idx - 1] = 0;
st = t, ed = s;
printf("%d", res - dinic());
return 0;
}