Blog of RuSun

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

P4001 [ICPC-Beijing 2006]狼抓兔子

P4001 [ICPC-Beijing 2006]狼抓兔子

正解应该是网格图的最小割转化为对偶图的最短路,这道题弧优化的 dinic 可以卡过。

对偶图最短路放在 P2046 [NOI2010] 海拔 里介绍。

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 = 1e6 + 10, M = 12e6 + 10;
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 find(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 = find(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 = find(st, INF))
res += flow;
return res;
}
void add(int x, int y, int z)
{
nxt[++idx] = hd[x];
hd[x] = idx;
edg[idx] = y;
wt[idx] = z;
}
int num(int x, int y)
{
return (x - 1) * m + y;
}
int main()
{
cin >> n >> m;
st = num(1, 1);
ed = num(n, m);
for (int i = st; i <= ed; i++)
hd[i] = -1;
for (int i = 1, a; i <= n; i++)
for (int j = 1; j < m; j++)
{
cin >> a;
int x = num(i, j),
y = num(i, j + 1);
add(x, y, a);
add(y, x, 0);
add(y, x, a);
add(y, x, 0);
}
for (int i = 1, a; i < n; i++)
for (int j = 1; j <= m; j++)
{
cin >> a;
int x = num(i, j),
y = num(i + 1, j);
add(x, y, a);
add(y, x, 0);
add(y, x, a);
add(x, y, 0);
}
for (int i = 1, a; i < n; i++)
for (int j = 1; j < m; j++)
{
cin >> a;
int x = num(i, j),
y = num(i + 1, j + 1);
add(x, y, a);
add(y, x, 0);
add(y, x, a);
add(x, y, 0);
}
cout << dinic();
return 0;
}