Blog of RuSun

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

P2153 [SDOI2009]晨跑

[P2153 [SDOI2009]晨跑]

每个点只能走一次,所以拆点。

Elaxia 耐力不太好,他希望在一个周期内跑的路程尽量短,但是又希望训练周期包含的天数尽量长。

考虑路程为费用,天数为流量,做费用流即可。

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 = 410, M = 5e4 + 10;
bool vis[N];
int n, m, st, ed, day, path, 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;
}
void ek()
{
while (spfa())
{
int t = incf[ed];
day += t;
path += t * d[ed];
for (int i = ed; i != st; i = edg[pre[i] ^ 1])
{
wt[pre[i]] -= t;
wt[pre[i] ^ 1] += t;
}
}
}
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;
st = 0;
ed = n << 1 | 1;
for (int i = st; i <= ed; i++)
hd[i] = -1;
add(st, 1, inf, 0);
add(1, st, 0, 0);
for (int i = 1, a, b, c; i <= m; i++)
{
cin >> a >> b >> c;
add(a, b + n, 1, c);
add(b + n, a, 0, -c);
}
for (int i = 2; i <= n; i++)
{
add(i + n, i, 1, 0);
add(i, i + n, 0, 0);
}
add(n << 1, ed, inf, 0);
add(ed, n << 1, 0, 0);
ek();
cout << day << ' ' << path;
return 0;
}