泛做题记录

• 伊基的故事 I - 道路重建
  • 网络流题，在跑完 dinic 后，在残量网络上找 \(f(u,v)=0\)，满足存在 \(s\rightarrow u,v \rightarrow t\) 的边。代码。
• [USACO05FEB] Secret Milking Machine G

  • 网络流题，明显二分答案。网络可以直接建反向边，因为流两次相当于没流。

    code

    const int N = 205;
    const int M = 8e4 + 5;
    const int inf = 2147483647;int head[N], ver[M], _edge[M], Next[M], tot = 1;
    int n, p, T, d[N], now[N], s, t;
    queue<int> que;
    int edge[M];void add(int u, int v, int w) {ver[++tot] = v, _edge[tot] = w;Next[tot] = head[u], head[u] = tot;
    }bool bfs() {memset(d, 0, sizeof d);while (que.size()) que.pop();d[s] = 1, que.push(s), now[s] = head[s];while (que.size()) {int u = que.front(); que.pop();for (int i = head[u]; i; i = Next[i]) {int v = ver[i], w = edge[i];if (w && !d[v]) {d[v] = d[u] + 1;now[v] = head[v];if (v == t) return 1;que.push(v);}}}return 0;
    }int dfs(int u, int f) {if (u == t || !f) return f;int res = 0;for (int &i = now[u]; i; i = Next[i]) {if (edge[i] && d[ver[i]] == d[u] + 1) {int k = dfs(ver[i], min(f, edge[i]));if (!k) d[ver[i]] = 0;edge[i] -= k, edge[i ^ 1] += k;f -= k, res += k;if (f <= 0) break;}}return res;
    }int dinic() {int res = 0, f;while (bfs()) while (f = dfs(s, inf)) res += f;return res;
    }bool check(int val) {LF(i, 2, tot) edge[i] = (_edge[i] <= val ? 1 : 0);if (dinic() >= T) return 1;return 0;
    }int main() {cin >> n >> p >> T;s = 1, t = n;LF(i, 1, p) {int u, v, w; cin >> u >> v >> w;add(u, v, w);add(v, u, w);}int l = 1, r = 1e6;while (l < r) {int mid = (l + r) >> 1;if (check(mid)) r = mid;else l = mid + 1;}cout << l;return 0;
    }