[TJOI 2018]智力竞赛

Description

题库链接

给出一张 \(m\) 个点的有向图。问可重最小路径覆盖是否 \(\leq n+1\) 。若不，求最多用 \(n+1\) 条路径去覆盖，最大化未覆盖点点权最小值。

\(1\leq n\leq 50,1\leq m\leq 500\)

Solution

\(\text{floyd}\) 传递闭包之后做最小路径覆盖...

若无解，考虑二分未覆盖点点权最小值，将点权小于 \(mid\) 的点加入图中做最小路径覆盖。

写这题其实是打 \(\text{dinic}\) 板子的...

Code

#include <bits/stdc++.h>
using namespace std;
const int N = 500+5, inf = ~0u>>1;

int n, m, G

, v
, k, u;
struct tt {int to, next, cap; }edge[N*N<<1];
int path[N<<1], top, S = (N<<1)-2, T = (N<<1)-1;
int sta[N<<1], cur[N<<1], dist[N<<1];
queue<int>Q;

bool bfs() {
memset(dist, -1, sizeof(dist));
Q.push(S); dist[S] = 1;
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = path[u], v; ~i; i = edge[i].next)
if (dist[v = edge[i].to] == -1 && edge[i].cap > 0)
Q.push(v), dist[v] = dist[u]+1;
}
return dist[T] != -1;
}
int dinic() {
int totflow = 0;
while (bfs()) {
int u = S; top = 0; memcpy(cur, path, sizeof(cur));
while (1) {
if (u == T) {
int loc, minflow = inf;
for (int i = 1; i <= top; i++) if (edge[sta[i]].cap < minflow) minflow = edge[sta[loc = i]].cap;
for (int i = 1; i <= top; i++) edge[sta[i]].cap -= minflow, edge[sta[i]^1].cap += minflow;
totflow += minflow; u = edge[sta[loc]^1].to, top = loc-1;
}
for (int &i = cur[u], v; ~i; i = edge[i].next)
if (dist[v = edge[i].to] == dist[u]+1 && edge[i].cap > 0) {
sta[++top] = i, u = v; break;
}
if (cur[u] == -1) {
if (top == 0) break;
dist[u] = -inf; u = edge[sta[top--]^1].to;
}
}
}
return totflow;
}
void add(int u, int v, int c) {
edge[++top] = (tt){v, path[u], c}; path[u] = top;
edge[++top] = (tt){u, path[v], 0}; path[v] = top;
}
bool judge(int mid) {
memset(path, top = -1, sizeof(path)); int cnt = 0;
for (int i = 1; i <= m; i++) add(S, i, 1), add(i+N-5, T, 1);
for (int i = 1; i <= m; i++)
if (v[i] < mid) {
++cnt;
for (int j = 1; j <= m; j++)
if (v[j] < mid && G[i][j])
add(i, j+N-5, 1);
}
return cnt-dinic() <= n+1;
}
void work() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= m; i++) {
scanf("%d%d", &v[i], &k);
for (int j = 1; j <= k; j++) scanf("%d", &u), G[i][u] = 1;
}
for (int k = 1; k <= m; k++)
for (int i = 1; i <= m; i++)
for (int j = 1; j <= m; j++)
G[i][j] |= (G[i][k]&G[k][j]);
if (judge(1000000000+1)) {puts("AK"); return; }
int L = 0, R = 1e9, ans;
while (L <= R) {
int mid = (L+R)>>1;
if (judge(mid)) ans = mid, L = mid+1;
else R = mid-1;
}
printf("%d\n", ans);
}
int main() {work(); return 0; }