Luogu P3425 [POI2005]KOS-Dicing

题目描述

https://www.luogu.com.cn/problem/P3425

Solution

我们考虑二分答案,然后利用最大流判断是否可行

建图:

$u$ 连 $t$,容量为 $x$

对于第 $i$ 局游戏,$s$ 连 $i$,容量为 $1$

$i$ 连 $u[i]$,容量为 $1$

$i$ 连 $v[i]$,容量为 $1$

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#include <iostream>
#include <cstdio>
#include <queue>
#define INF 1000000000
#define maxn 20010
using namespace std;

int n, m, u[maxn], v[maxn];

struct Edge {
int to, next, w;
} e[1000000]; int c1, head[maxn];
inline void add_edge(int u, int v, int w) {
e[c1].to = v; e[c1].w = w;
e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w) {
add_edge(u, v, w); add_edge(v, u, 0);
}

int dep[maxn], s, t;
bool bfs() {
fill(dep, dep + maxn, 0); dep[s] = 1;
queue<int> Q; Q.push(s);
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w;
if (w > 0 && !dep[v]) {
dep[v] = dep[u] + 1;
Q.push(v); if (v == t) return 1;
}
}
}
return 0;
}

int dfs(int u, int _w) {
if (u == t || !_w) return _w;
int s = 0;
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w;
if (w > 0 && dep[v] == dep[u] + 1) {
int di = dfs(v, min(_w - s, w));
e[i].w -= di; e[i ^ 1].w += di;
s += di; if (s == _w) break;
}
}
if (s < _w) dep[u] = 0;
return s;
}

int dinic() {
int mf = 0;
while (bfs()) mf += dfs(s, INF);
return mf;
}

void init() {
fill(head, head + maxn, -1); c1 = 0;
}

bool check(int x) { init();
for (int i = 1; i <= n; ++i) Add_edge(i, t, x);
for (int i = 1; i <= m; ++i) {
Add_edge(s, i + n, 1);
Add_edge(i + n, u[i], 1);
Add_edge(i + n, v[i], 1);
}
return dinic() == m;
}

int ans[maxn];
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);

cin >> n >> m; s = 0; t = n + m + 1;
for (int i = 1; i <= m; ++i) cin >> u[i] >> v[i];
int l = 1, r = m, mid, ans = 0;
while (l <= r) {
mid = l + r >> 1;
if (check(mid)) ans = mid, r = mid - 1;
else l = mid + 1;
} check(ans);
for (int u = 1; u <= m; ++u)
for (int i = head[u + n]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w;
if (v && !w) ::ans[u] = v;
}
cout << ans << "\n";
for (int i = 1; i <= m; ++i) cout << (::ans[i] == u[i]) << "\n";
return 0;
}