算是网络最大流的模板题了,虽然匈牙利算法也能AC,而且在这道题上要更简单(比如可以直接输出最佳飞行员配对方案),但毕竟这套题叫网络流24题。
没什么好说的,直接上代码了。
代码实现如下:
#include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (register int i = a; i <= b; i++) const int inf = 0x3f3f3f3f, maxn = 1e5 + 5; int n, m, S, T, num_edge = 1, ans = 0; int cur[maxn], dep[maxn], head[maxn]; queue<int> q; struct node{ int to, nxt, dis; }edge[maxn]; void origin(){memset(head, -1, sizeof(head));} int read() { int x = 0, flag = 0; char ch = ‘ ‘; while (ch != ‘-‘ && (ch < ‘0‘ || ch > ‘9‘)) ch = getchar(); if (ch == ‘-‘) { flag = 1; ch = getchar(); } while (ch >= ‘0‘ && ch <= ‘9‘) { x = x * 10 + ch - ‘0‘; ch = getchar(); } return flag ? -x : x; } void addedge(int from, int to, int dis) { edge[++num_edge].nxt = head[from]; edge[num_edge].to = to; edge[num_edge].dis = dis; head[from] = num_edge; } int bfs() { memset(dep, 0, sizeof(dep)); while (!q.empty()) q.pop(); rep(i, 0, n + 1) cur[i] = head[i]; dep[S] = 1; q.push(S); while (!q.empty()) { int u = q.front(); q.pop(); for (int i = head[u]; ~i; i = edge[i].nxt) { int v = edge[i].to; if (!dep[v] && edge[i].dis) { dep[v] = dep[u] + 1; q.push(v); } } } if (dep[T]) return 1; return 0; } int dfs(int u, int flow) { if (u == T || !flow) return flow; int d, used = 0; for (int i = cur[u]; ~i; i = edge[i].nxt) { cur[u] = i; int v = edge[i].to; if (dep[v] == dep[u] + 1 && (d = dfs(v, min(edge[i].dis, flow)))) { used += d; flow -= d; edge[i].dis -= d; edge[i ^ 1].dis += d; if (!flow) break; } } if (!used) dep[u] = -2; return used; } int dinic() { int ans = 0; while (bfs()) ans += dfs(S, inf); return ans; } void write(int x) { if (x < 0) { putchar(‘-‘); x = -x; } if (x > 9) write(x / 10); putchar(x % 10 + ‘0‘); } int main() { origin(); m = read(), n = read(); S = 0, T = n + 1; while (1) { int u, v; u = read(), v = read(); if (u == -1 && v == -1) break; addedge(u, v, inf); addedge(v, u, 0); } rep(i, 1, m) { addedge(S, i, 1); addedge(i, S, 0); } rep(i, m + 1, n) { addedge(i, T, 1); addedge(T, i, 0); } ans = dinic(); if (!ans) { printf("No Solution!"); return 0; } write(ans); printf("\n"); for (int i = 2; i <= num_edge; i += 2) { if (edge[i].to != S && edge[i ^ 1].to != S) if (edge[i].to != T && edge[i ^ 1].to != T) if (edge[i ^ 1].dis) { write(edge[i ^ 1].to); printf(" "); write(edge[i].to); printf("\n"); } } return 0; }
原文:https://www.cnblogs.com/Kirisame-Marisa/p/10392615.html