Blog of RuSun

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

P4014 分配问题

P4014 分配问题

裸的带权匹配,分别求最大值和最小值(最长路、最短路),为了简便,我们求最大值的时候将边权变为它的相反数,求最小值的相反数即答案。

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 = 110, M = 3e4 + 10;
bool vis[N];
int n, st, ed, d[N], pre[N], incf[N];
int idx = -1, hd[N], nxt[M], edg[M], wt[M], f[M];
bool spfa ()
{
for (int i = st; i <= ed; i++)
incf[i] = 0;
for (int i = st; i <= ed; i++)
d[i] = INF;
incf[st] = INF;
d[st] = 0;
queue <int> q;
q.push(st);
vis[st] = true;
while (!q.empty())
{
int t = q.front();
q.pop();
vis[t] = false;
for (int i = hd[t]; ~i; i = nxt[i])
if (d[t] + f[i] < d[edg[i]] && wt[i])
{
d[edg[i]] = d[t] + f[i];
pre[edg[i]] = i;
incf[edg[i]] = min(wt[i], incf[t]);
if (!vis[edg[i]])
{
vis[edg[i]] = true;
q.push(edg[i]);
}
}
}
return incf[ed] > 0;
}
int ek ()
{
int res = 0;
while (spfa())
{
int t = incf[ed];
res += t * d[ed];
for (int i = ed; i != st; i = edg[pre[i] ^ 1])
{
wt[pre[i]] -= t;
wt[pre[i] ^ 1] += t;
}
}
return res;
}
void add (int a, int b, int c, int d)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
f[idx] = d;
}
int main ()
{
cin >> n;
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, 0);
add(i, st, 0, 0);
}
for (int i = 1, a; i <= n; i++)
for (int j = n + 1; j < ed; j++)
{
cin >> a;
add(i, j, 1, a);
add(j, i, 0, -a);
}
for (int i = n + 1; i < ed; i++)
{
add(i, ed, 1, 0);
add(ed, i, 0, 0);
}
cout << ek() << endl;
for (int i = 0; i <= idx; i += 2)
{
wt[i] += wt[i ^ 1];
wt[i ^ 1] = 0;
f[i] = -f[i];
f[i ^ 1] = -f[i ^ 1];
}
cout << -ek();
return 0;
}