Blog of RuSun

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

P2045 方格取数加强版

P2045 方格取数加强版

拆点,每个格子中流量为 $1$ 。

查看代码
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#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <climits>
#define INF INT_MAX
using namespace std;
const int N = 5e3 + 10, M = 2e4 + 10;
bool vis[N];
int n, k, 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 = 0; i <= ed; i++)
d[i] = -INF;
d[st] = 0;
for (int i = 0; i <= ed; i++)
incf[i] = 0;
incf[st] = INF;
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(incf[t], wt[i]);
if (!vis[edg[i]])
{
q.push(edg[i]);
vis[edg[i]] = true;
}
}
}
return incf[ed] > 0;
}
int ek()
{
int res = 0;
while (spfa())
{
int t = incf[ed];
res += d[ed] * t;
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 num(int x, int y)
{
if (x < 1 || x > n || y < 1 || y > n)
return -1;
return (x - 1) * n + y - 1;
}
int main()
{
cin >> n >> k;
st = n * n * 2 + 1;
ed = n * n * 2 + 2;
for (int i = 0; i <= ed; i++)
hd[i] = -1;
add(st, num(1, 1) << 1, k, 0);
add(num(1, 1) << 1, st, 0, 0);
for (int i = 1, a; i <= n; i++)
for (int j = 1; j <= n; j++)
{
int t = num(i, j);
cin >> a;
add(t << 1, t << 1 | 1, 1, a);
add(t << 1 | 1, t << 1, 0, -a);
add(t << 1, t << 1 | 1, INF, 0);
add(t << 1 | 1, t << 1, 0, 0);
int h = num(i, j + 1);
if (~h)
{
add(t << 1 | 1, h << 1, INF, 0);
add(h << 1, t << 1 | 1, 0, 0);
}
h = num(i + 1, j);
if (~h)
{
add(t << 1 | 1, h << 1, INF, 0);
add(h << 1, t << 1 | 1, 0, 0);
}
}
add(num(n, n) << 1 | 1, ed, k, 0);
add(ed, num(n, n) << 1 | 1, 0, 0);
cout << ek();
return 0;
}