Blog of RuSun

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

P2517 [HAOI2010]订货

P2517 [HAOI2010]订货

利用费用流的最大流性质,一个点到汇点的流量即需求量,从源点来的流量为购买,从上一个点来的流量为存贮。

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 = 60, M = 310;
bool vis[N];
int n, m, S, 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++)
d[i] = inf;
for (int i = st; i <= ed; i++)
incf[i] = 0;
d[st] = 0;
queue <int> q;
q.push(st);
vis[st] = true;
incf[st] = inf;
while (!q.empty())
{
int t = q.front();
q.pop();
vis[t] = false;
for (int i = hd[t]; ~i; i = nxt[i])
if (wt[i] && d[t] + f[i] < d[edg[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]] = false;
}
}
}
return incf[ed] > 0;
}
int ek()
{
int res = 0;
while (spfa())
{
int t = incf[ed];
res += d[ed] * incf[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 >> m >> S;
st = 0;
ed = n + 1;
for (int i = st; i <= ed; i++)
hd[i] = -1;
for (int i = 1, a; i <= n; i++)
{
cin >> a;
add(i, ed, a, 0);
add(ed, i, 0, 0);
if (i < n)
{
add(i, i + 1, S, m);
add(i + 1, i, 0, -m);
}
}
for (int i = 1, a; i <= n; i++)
{
cin >> a;
add(st, i, inf, a);
add(i, st, 0, -a);
}
cout << ek();
return 0;
}