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P2756 飞行员配对方案问题

P2756 飞行员配对方案问题

做了几道蓝色的水题

题面

题目链接

解法

二分图网络流板子,用来强迫自己敲一遍$dinic$。

AC代码

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/**
* @file: P2756.cpp
* @author: yaoxi-std
* @url: https://www.luogu.com.cn/problem/P2756
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp &x) {
bool sign = false;
char ch = getchar();
long double tmp = 1;
for (; !isdigit(ch); ch = getchar())
sign |= (ch == '-');
for (x = 0; isdigit(ch); ch = getchar())
x = x * 10 + (ch ^ 48);
if (ch == '.')
for (ch = getchar(); isdigit(ch); ch = getchar())
tmp /= 10.0, x += tmp * (ch ^ 48);
return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
write(x / 10);
putchar((x % 10) ^ 48);
}
const int MAXN = 350;
const int MAXM = 1e5 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct dinic {
struct edges {
int v, w, next;
} edge[MAXM];
int n, tot, head[MAXN];
int cur[MAXN], lev[MAXN];
void init(int n) {
this->n = n, tot = 0;
fill(head, head + n + 1, -1);
}
void addedge(int u, int v, int w) {
edge[tot] = {v, w, head[u]};
head[u] = tot++;
edge[tot] = {u, 0, head[v]};
head[v] = tot++;
}
bool bfs(int s, int t) {
fill(lev, lev + n + 1, -1);
queue<int> que;
que.push(s), lev[s] = 0;
while (!que.empty()) {
int u = que.front();
que.pop();
for (int i = head[u]; ~i; i = edge[i].next) {
if (edge[i].w && lev[edge[i].v] == -1) {
lev[edge[i].v] = lev[u] + 1;
que.push(edge[i].v);
}
}
}
return lev[t] != -1;
}
int dfs(int u, int t, int mx) {
if (u == t || mx == 0)
return mx;
int ret = 0;
for (int &i = cur[u]; ~i; i = edge[i].next) {
int v = edge[i].v, w = edge[i].w;
if (lev[v] != lev[u] + 1)
continue;
int tmp = dfs(v, t, min(mx, w));
edge[i].w -= tmp, edge[i ^ 1].w += tmp;
mx -= tmp, ret += tmp;
if (mx == 0)
break;
}
return ret;
}
int maxflow(int s, int t) {
int ret = 0;
while (bfs(s, t)) {
copy(head, head + n + 1, cur);
ret += dfs(s, t, INF);
}
return ret;
}
};
int n, m, s, t;
dinic nf;
signed main() {
read(m), read(n);
s = n + 1, t = n + 2;
nf.init(n + 2);
for (int i = 1; i <= m; ++i)
nf.addedge(s, i, 1);
for (int i = m + 1; i <= n; ++i)
nf.addedge(i, t, 1);
int u, v;
while (scanf("%lld%lld", &u, &v), u != -1 && v != -1)
nf.addedge(u, v, INF);
write(nf.maxflow(s, t)), putchar('\n');
for (int i = 0; i < nf.tot; i += 2)
if (nf.edge[i].w == INF - 1)
printf("%lld %lld\n", nf.edge[i ^ 1].v, nf.edge[i].v);
return 0;
}