題目: UVa - 11380 - Down Went The Titanic
題目說明
給定一個圖,上面有薄冰 ‘.’ 或 ‘*‘, 厚冰 ‘@’, 木塊 ‘#’,一開始人都在 ‘*‘ 上,薄冰只能走一次就會沉掉,厚冰次數不限,如果人走到木塊上就獲救了,但是一個木塊的容量只有 p,求最多能有多少人獲救。
UVA 11380 - Down Went The Titanic(网络流)_Remilia’s-CSDN博客
解題思路
最大流,依照上面的說明建邊後使用 Dinic 求解即可。
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 = 2005;
int S, T; int rn[MXN][MXN]; int l[MXN];
int Y, X, P; int tot; int d[] = { 1, 0, -1, 0, 1 };
void init() { CLR(rn); }
void read() { tot = Y * X; T = 2 * tot + 1;
char c; FOR(y, 0, Y) FOR(x, 0, X) { int u = X * y + x + 1; cin >> c; switch (c) { case '*': rn[u][u + tot] = 1; rn[S][u] = 1; break;
case '.': rn[u][u + tot] = 1; break;
case '@': rn[u][u + tot] = INF; break;
case '#': rn[u][u + tot] = INF; rn[u + tot][T] = P; break; }
FOR(i, 0, 4) { int dx = x + d[i]; int dy = y + d[i + 1]; if (dx < 0 || dx >= X || dy < 0 || dy >= Y) continue;
int v = X * dy + dx + 1; rn[u + tot][v] = INF; } } }
bool dinicBFS() { CLR(l);
queue<int> q; q.push(S); l[S] = 1;
while (!q.empty()) { int u = QPOP(q);
FOR(v, 0, T + 1) { if (l[v] || !rn[u][v]) continue; l[v] = l[u] + 1; q.push(v); } }
return l[T]; }
int dinicDFS(int u, int cp) { if (u == T) return cp;
int tmp = cp;
FOR(v, 0, T + 1 && tmp) { if (l[v] != l[u] + 1 || !rn[u][v]) continue; int bt = dinicDFS(v, min(rn[u][v], tmp)); rn[u][v] -= bt; rn[v][u] += bt; tmp -= bt; }
return cp - tmp; }
int maximumFlow() { int mf = 0; while (dinicBFS()) mf += dinicDFS(S, INF); return mf; }
void solve() { cout << maximumFlow() << '\n'; }
int main() { while (cin >> Y >> X >> P) { init(); read(); solve(); } }
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參考資料
UVA 11380 - Down Went The Titanic(网络流)_Remilia’s-CSDN博客