[TJOI 2018]智力竞赛

Posted 2020-12-04 navi-awson

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。

(1leq nleq 50,1leq mleq 500)

Solution

( ext{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 ( ext{dinic}) 板子的...

Code

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

int n, m, G[N][N], v[N], 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
", ans);
}
int main() {work(); return 0; }