Blog of RuSun

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

P1361 小M的作物

P1361 小M的作物

这类问题中,最小割的意义是最小放弃多少价值。两种物品同时选择一方则可以在原来的基础上增加利益,一般新建一个点,从该方向此点连接价值为新增价值的边,在从此点向两种物品连接 $inf$ 的边,表示不可割。那么,只有在两点同时割掉另一方时,此边才可以不被割,即不被放弃。

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 = 3e3 + 10, M = 3e6 + 10;
int n, m, st, ed, tot, d[N], cur[N];
int idx = -1, hd[N], nxt[M], edg[M], wt[M];
bool bfs()
{
memset(d, -1, sizeof d);
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 main()
{
memset(hd, -1, sizeof - 1);
cin >> n;
st = 0;
ed = n + 1;
for (int i = 1, a; i <= n; i++)
{
cin >> a;
tot += a;
add(st, i, a);
add(i, st, 0);
}
for (int i = 1, a; i <= n; i++)
{
cin >> a;
tot += a;
add(i, ed, a);
add(ed, i, 0);
}
cin >> m;
for (int i = n + 1 + 1, k, c1, c2, a; i <= n + 1 + m * 2; i += 2)
{
cin >> k >> c1 >> c2;
tot += c1 + c2;
add(st, i, c1);
add(i, st, 0);
add(i + 1, ed, c2);
add(ed, i + 1, 0);
for (int j = 1; j <= k; j++)
{
cin >> a;
add(i, a, INF);
add(a, i, 0);
add(a, i + 1, INF);
add(i + 1, a, 0);
}
}
cout << tot - dinic();
return 0;
}