【定义】
【最大流】
从源点向连边流出流量 fi ,总计为 f，在到达汇点时，对每条边的流量限制ei都有，fi<ci
令 f 尽量大，这个 f 被称为最大流
【最小割】
有图 V，给出点 s，t，去掉一条边的代价为其流量限制，求使 s 无法到 t 的最小代价
这个代价被称为最小割
经过一些我看不懂（我会继续看的QAQ）证明，我们可以得到结论
【结论】 最大流等于最小割
【方法】
【Ford】
通过增加反向流量的方法，提供一种“反悔“的操作，从而支持流量从源点到汇点的改变和增加
新增加的路径被称为增广路
记得当时学匈牙利算法时候看过
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> using namespace std; const int MAXN = 10010; const int MAXM = 200010; const int INF = (1 << 31) - 1; int n, m, s, t; struct V { int to; int rev; int cap; int nt; }edge[MAXM]; int st[MAXN], top=0; int vis[MAXN]; void add(int x, int y, int val) { top++; edge[top] = { y, top + 1, val, st[x] }; st[x] = top; top++; edge[top] = { x, top - 1, 0 , st[y] }; st[y] = top; } int dfs(int num, int srm) { if (num == t) return srm; vis[num] = 1; for (int i = st[num]; i!=-1; i = edge[i].nt) { int to = edge[i].to; if (!vis[to] && edge[i].cap > 0) { int minn = min(srm, edge[i].cap); int d = dfs(to, minn); if (d <= 0) continue; edge[i].cap -= d; edge[edge[i].rev].cap += d; return d; } } return 0; } int max_flow() { int flow = 0; while (1) { memset(vis, 0, sizeof(vis)); int d = dfs(s, INF); if (!d) break; flow += d; } return flow; } int main() { scanf("%d%d%d%d", &n, &m, &s, &t); memset(st, -1, sizeof(st)); for (int i = 1; i <= m; i++) { int x, y, z; scanf("%d%d%d", &x, &y, &z); add(x, y, z); } printf("%d", max_flow()); return 0; }
【dinic】
通过对节点进行分层的操作，从而使增广路的效率提高（使其不会向回运动）
同时，对走过的路进行标记（或者取消），通过节省路径数，提高效率
#include<cstdio> #include<queue> #include<cstring> #define ll long long using namespace std; const int MAXN = 10010; const int MAXM = 200010; const ll INF = (1ll << 62) - 1; struct note { int to; int nt; int rev; ll cal; }; struct edge { note arr[MAXM]; int st[MAXN]; int dis[MAXN]; int cur[MAXN]; int depth[MAXN]; int top; int n, m, s, t; edge() { memset(st, -1, sizeof(st)); memset(depth, -1, sizeof(depth)); memset(dis, -1, sizeof(dis)); top = 0; } void read() { top = 0; scanf("%d%d%d%d", &n, &m, &s, &t); for (int i = 1; i <= m; i++) { int x, y; ll z; scanf("%d%d%lld", &x, &y, &z); add(x, y, z); } } bool dep() { queue<int> q; q.push(s); memset(depth, -1, sizeof(depth)); depth[s] = 0; while (!q.empty()) { int v = q.front(); q.pop(); for (int i = st[v]; i != -1; i = arr[i].nt) { int to = arr[i].to; if (!arr[i].cal) continue; if (depth[to] != -1) continue; depth[to] = depth[v] + 1; q.push(to); } } return (depth[t] != -1); } void add(int x, int y, ll z) { top++; arr[top] = { y,st[x],top + 1,z }; st[x] = top; top++; arr[top] = { x,st[y],top - 1,0 }; st[y] = top; } ll min_dis() { } ll dfs(int now, ll val) { if (now == t || !val) return val; ll flow = 0; for (int& i = cur[now]; i != -1; i = arr[i].nt) { int to = arr[i].to; if (depth[to] != depth[now] + 1) continue; ll f = dfs(to, min(arr[i].cal, val)); if (!f || !arr[i].cal) continue; flow += f; arr[i].cal -= f; arr[arr[i].rev].cal += f; val -= f; if (!val) return flow; } return flow; } ll dinic() { ll flow = 0; ll f; while (dep()) { for (int i = 1; i <= n; i++) cur[i] = st[i]; while (f = dfs(s, INF)) flow += f; } return flow; } }; edge road; int main() { int T; road.read(); printf("%lld", road.dinic()); return 0; }