POJ 3204 Ikki's Story I - Road Reconstruction（最大流）

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POJ 3204 Ikki‘s Story I - Road Reconstruction

題目連結

題意：給定一個有向圖，求出最大流後，問哪些邊增加容量後，可以使最大流增加

思路：對於一個可以增加的，必然原來就是滿流，並且從源點到匯點，的一條路徑上，都是還有殘留容量的，這樣只要從源點和匯點分別出發dfs一遍，標記掉經過點，然後枚舉滿流邊，如果兩端都是標記過的點，這個邊就是可以增加的

代碼：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int MAXNODE = 505;const int MAXEDGE = 100005;typedef int Type;const Type INF = 0x3f3f3f3f;struct Edge {int u, v;Type cap, flow;int cnt;Edge() {}Edge(int u, int v, Type cap, Type flow) {this->u = u;this->v = v;this->cap = cap;this->flow = flow;this->cnt = 0;}};struct Dinic {int n, m, s, t;Edge edges[MAXEDGE];int first[MAXNODE];int next[MAXEDGE];bool vis[MAXNODE];Type d[MAXNODE];int cur[MAXNODE];vector<int> cut;void init(int n) {this->n = n;memset(first, -1, sizeof(first));m = 0;}void add_Edge(int u, int v, Type cap) {edges[m] = Edge(u, v, cap, 0);next[m] = first[u];first[u] = m++;edges[m] = Edge(v, u, 0, 0);next[m] = first[v];first[v] = m++;}bool bfs() {memset(vis, false, sizeof(vis));queue<int> Q;Q.push(s);d[s] = 0;vis[s] = true;while (!Q.empty()) {int u = Q.front(); Q.pop();for (int i = first[u]; i != -1; i = next[i]) {Edge& e = edges[i];if (!vis[e.v] && e.cap > e.flow) {vis[e.v] = true;d[e.v] = d[u] + 1;Q.push(e.v);}}}return vis[t];}Type dfs(int u, Type a) {if (u == t || a == 0) return a;Type flow = 0, f;for (int &i = cur[u]; i != -1; i = next[i]) {Edge& e = edges[i];if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {e.flow += f;edges[i^1].flow -= f;flow += f;a -= f;if (a == 0) break;}}return flow;}Type Maxflow(int s, int t) {this->s = s; this->t = t;Type flow = 0;while (bfs()) {for (int i = 0; i < n; i++)cur[i] = first[i];flow += dfs(s, INF);}return flow;}void MinCut() {cut.clear();for (int i = 0; i < m; i += 2) {if (vis[edges[i].u] && !vis[edges[i].v])cut.push_back(i);}}bool mark[MAXNODE][2];void find(int u, int tp) {mark[u][tp] = true;for (int i = first[u]; i + 1; i = next[i]) {int v = edges[i].v;if (mark[v][tp]) continue;if (tp == 0 && i % 2) continue;if (tp && i % 2 == 0) continue;if (edges[i^tp].cap == edges[i^tp].flow) continue;find(v, tp);}}int solve() {Maxflow(0, n - 1);memset(mark, false, sizeof(mark));find(0, 0);find(n - 1, 1);int ans = 0;for (int i = 0; i < m; i += 2)if (edges[i].cap == edges[i].flow && mark[edges[i].u][0] && mark[edges[i].v][1]) ans++;return ans;}} gao;int n, m;int main() {while (~scanf("%d%d", &n, &m)) {gao.init(n);int u, v, w;while (m--) {scanf("%d%d%d", &u, &v, &w);gao.add_Edge(u, v, w);}printf("%d\n", gao.solve());}return 0;}

POJ 3204 Ikki's Story I - Road Reconstruction（最大流）