Blog of RuSun

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

P2764 最小路径覆盖问题

P2764 最小路径覆盖问题

最坏的情况是每个路径只用一个点,这是一个可行解。考虑如何让答案更优,对于一条边,如果两个端点中起点没有连接路径作为出点,终点没有连接路径作为入点,那么可以通过这条边将两个端点的路径合并,答案更优,然后起点作为出点不能再用了,终点作为入点不能再用了。于是考虑拆点后,二分图的匹配。

考虑输出方案,对于每一个没有访问的点,向前后寻找是否有可以构成的路径的点,依次递归输出即可。

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 = 310, M = 2e4 + 10;
bool vis[N];
int n, m, st, ed, res, root, 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;
}
void print_forward(int x)
{
vis[x] = true;
if (x != root)
cout << x << ' ';
for (int i = hd[x]; ~i; i = nxt[i])
if (edg[i] != st && !wt[i])
print_forward(edg[i] - n);
}
void print_back(int x)
{
vis[x] = true;
for (int i = hd[x + n]; ~i; i = nxt[i])
if (edg[i] != ed && wt[i])
print_back(edg[i]);
if (x != root)
cout << x << ' ';
}
int main()
{
cin >> n >> m;
st = 0;
ed = n << 1 | 1;
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, a, b; i <= m; i++)
{
cin >> a >> b;
b += n;
add(a, b, 1);
add(b, a, 0);
}
for (int i = n + 1; i < ed; i++)
{
add(i, ed, 1);
add(ed, i, 0);
}
res = dinic();
for (int i = 1; i <= n; i++)
if (!vis[i])
{
root = i;
print_back(i);
cout << i << ' ';
print_forward(i);
cout << endl;
}
cout << n - res;
return 0;
}