Blog of RuSun

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

P2294 [HNOI2005]狡猾的商人

P2294 [HNOI2005]狡猾的商人

直接差分约束即可,找负环判断是否有解。

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 = 110, M = 3e3 + 10;
bool vis[N];
int n, m, st, d[N], cnt[N];
int idx, hd[N], nxt[M], edg[M], wt[M];
bool spfa ()
{
for (int i = 0; i <= n + 1; i++)
vis[i] = false;
for (int i = 0; i <= n + 1; i++)
d[i] = -INF;
for (int i = 0; i <= n + 1; i++)
cnt[i] = 0;
queue <int> q;
d[st] = 0;
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] + wt[i] > d[edg[i]])
{
d[edg[i]] = d[t] + wt[i];
cnt[edg[i]] = cnt[t] + 1;
if (cnt[edg[i]] > n + 1)
return false;
if (!vis[edg[i]])
{
q.push(edg[i]);
vis[edg[i]] = true;
}
}
}
return true;
}
void add (int a, int b, int c)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
}
int main ()
{
int T;
cin >> T;
while (T--)
{
cin >> n >> m;
idx = -1;
st = n + 1;
for (int i = 0; i <= n + 1; i++)
hd[i] = -1;
for (int i = 0; i <= n; i++)
add(st, i, 0);
for (int i = 1, a, b, c; i <= m; i++)
{
cin >> a >> b >> c;
add(a - 1, b, c);
add(b, a - 1, -c);
}
if (spfa())
cout << "true" << endl;
else
cout << "false" << endl;
}
return 0;
}