題目: UVa - 10806 - Dijkstra, Dijkstra.

題目說明

給你一個無向圖,每條邊有 cost,現在問如果不走已經走過的邊,則從 S 到 T,然後再從 T 到 S 的最少 cost 是多少,如果無法達成則輸出 “Back to jail”。

Programming學習筆記: UVa 10806 Dijkstra, Dijkstra

解題思路

建圖後求 MCMF 即可,由於是雙向圖,所以將路線的邊 Capacity 都設為 1,源點到監獄及終點到匯點的 Capacity 設為 2 即可。

參考解法

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#include <bits/stdc++.h>

using namespace std;

#define FOR(i, a, b) for (int i = a; i < b; ++i)
#define CLR(c) memset(c, 0, sizeof(c))

static auto __ = []
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
return 0;
}();

template <typename T>
T QPOP(queue<T>& q)
{
T tmp = q.front();
q.pop();
return tmp;
}

const int INF = (int)1e9;
const int MXN = 105;

vector<int> e[MXN];
int c[MXN][MXN];
int f[MXN][MXN];
int cost[MXN][MXN];
int d[MXN];
int p[MXN];
bool inQ[MXN];

int S, T;
int N, M;

void addEdge(int u, int v, int Cap = 0, int Cost = 0)
{
e[u].push_back(v);
c[u][v] = Cap;
cost[u][v] = Cost;
}

void init()
{
T = N + 1;
for (auto& v : e) v.clear();
CLR(f);
}

void read()
{
cin >> M;

int u, v, C;
while (M--)
{
cin >> u >> v >> C;
addEdge(u, v, 1, C);
addEdge(v, u, 1, C);
}

addEdge(S, 1, 2, 0);
addEdge(1, S, 2, 0);

addEdge(N, T, 2, 0);
addEdge(T, N, 2, 0);
}

bool SPFA()
{
fill(d, d + MXN, INF);
d[S] = 0;

CLR(inQ);
queue<int> q;

q.push(S);
inQ[S] = true;

while (!q.empty())
{
auto u = QPOP(q);
inQ[u] = false;

for (auto& v : e[u])
{
if (c[u][v] > f[u][v] && d[u] + cost[u][v] < d[v])
{
d[v] = d[u] + cost[u][v];
p[v] = u;
if (!inQ[v]) q.push(v), inQ[v] = true;
}
if (f[v][u] > 0 && d[u] + (-cost[v][u]) < d[v])
{
d[v] = d[u] + (-cost[v][u]);
p[v] = u;
if (!inQ[v]) q.push(v), inQ[v] = true;
}
}
}

return d[T] != INF;
}

int augment(int u, int v, int bottleNeck)
{
if (v == S) return bottleNeck;
bottleNeck = augment(p[u], u, min(c[u][v] - f[u][v], bottleNeck));
f[u][v] += bottleNeck;
f[v][u] -= bottleNeck;
return bottleNeck;
}

int MCMF()
{
int mnC = 0, mf = 0;

while (SPFA())
{
mnC += d[T];
mf += augment(p[T], T, INF);
if (mf >= 2) break;
}

if (mf < 2) return -1;
return mnC;
}

void solve()
{
auto mnC = MCMF();
if (mnC != -1) cout << mnC << '\n';
else cout << "Back to jail\n";
}

int main()
{
while (cin >> N && N)
{
init();
read();
solve();
}
}

參考資料

Programming學習筆記: UVa 10806 Dijkstra, Dijkstra