车站分级

今日份DAG呈上

题目

https://www.acwing.com/problem/content/458/

思路

题意：同一趟车次内，停靠的车站\(a\)的等级严格大于未停靠的车站\(b\)的等级

所以可以根据\(a>b\)来建边（即，所有未停靠站建边指向所有停靠站）

优化：对于两个点集之间，可以在中间建立一个虚拟源点，复杂度O(n*m)变成O(n+m)

把虚拟源点的值取为左边点集的最大值，左集连到虚拟源点的权值为0，点到右集的权值为1

每条车路径对应一个虚拟源点\(n+i\)

Key：拓扑排序+虚拟源点

Code

#include <bits/stdc++.h>

using namespace std;
const int N = 2005, M = N * N; //外加1000个虚拟点

int n, m, cnt = 1;
int h[N], e[M], ne[M], w[M], idx;
int d[N], a[N], dis[N];
bool stop[N];

void add (int a, int b, int c) {
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++;
    d[b] ++; //topsort特供，记得入度
}

void topsort () {
    queue <int> q;
    for (int i = 1; i <= n + m; i ++) //n + m
        if (d[i] == 0)
            q.push (i);

    while (!q.empty()) {
        int t = q.front();
        q.pop();
        a[cnt ++] = t;

        for (int i = h[t]; i != -1; i = ne[i]) {
            int j = e[i];
            d[j] --;
            if (d[j] == 0)
                q.push (j);
        }
    }
          
}

int main () {
    memset (h, -1, sizeof h);
    cin >> n >> m;
    for (int i = 1; i <= m; i ++) {
        memset (stop, false, sizeof stop);
        int k;
        cin >> k;
        int st = n, ed = 1;
        for (int j = 0; j < k; j ++) {
            int s;
            cin >> s;
            st = min (st, s), ed = max (ed, s);
            stop[s] = true; //代表该点是站点
        }

        //重点来了，建图要用虚拟源点
        int virtue_n = n + i; //是 i 
        for (int j = st; j <= ed; j ++) {
            if (stop[j]) //下标写错痛哭一万年
                add (virtue_n, j, 1);
            else
                add (j, virtue_n, 0);
        }
        
    }

    topsort();

    for (int i = 1; i <= n; i ++)
        dis[i] = 1;

    for (int i = 0; i < n + m; i ++) {
        int j = a[i];
        for (int k = h[j]; k != -1; k = ne[k]) {
            int t = e[k];
            dis[t] = max (dis[t], dis[j] + w[k]);
        }
    }

    int ans = 0;
    for (int i = 1; i <= n; i ++)
        ans = max (ans, dis[i]);
    cout << ans << endl;

}

BB

提高课77 AC 祭

标签：idx,int,ed,拓扑,源点,456,st,虚拟,AcWing
来源： https://www.cnblogs.com/CTing/p/16201096.html