解题报告 『酒店之王（网络最大流 + 拆点）』

Posted 2021-02-15 kirisame-marisa

原题地址

网络流板子题 + 拆点，个人觉得蓝题比较合适。

尽管我一开始只得了10分。

具体还是看代码吧。

 

代码实现如下：

#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (register int i = a; i <= b; i++)

const int inf = 0x3f3f3f3f, maxn = 1e4 + 5;

int n, p, q, S, T, ans = 0, num_edge = -1;
int cur[maxn], dep[maxn], head[maxn];

queue<int> Q;

struct node{
    int to, dis, nxt;
}edge[maxn];

void origin(){memset(head, -1, sizeof(head));}

int read()
{
    int x = 0, flag = 0;
    char ch = ‘ ‘;
    while (ch != ‘-‘ && (ch < ‘0‘ || ch > ‘9‘)) ch = getchar();
    if (ch == ‘-‘)
    {
        flag = 1;
        ch = getchar();
    }
    while (ch >= ‘0‘ && ch <= ‘9‘)
    {
        x = (x << 1) + (x << 3) + ch - ‘0‘;
        ch = getchar();
    }
    return flag ? -x : x;
}

void addedge(int from, int to, int dis)
{
    edge[++num_edge].nxt = head[from];
    edge[num_edge].to = to;
    edge[num_edge].dis = dis;
    head[from] = num_edge;
}

int bfs(int S, int T)
{
    memset(dep, 0, sizeof(dep));
    while (!Q.empty()) Q.pop();
    memcpy(cur, head, sizeof(head));//一开始用的for循环，T了9个点，改成memcpy后就过了. 
    dep[S] = 1;
    Q.push(S);
    while (!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        for (int i = head[u]; ~i; i = edge[i].nxt)
        {
            int v = edge[i].to;
            if (!dep[v] && edge[i].dis)
            {
                dep[v] = dep[u] + 1;
                Q.push(v);
            }
        }
    }
    if (dep[T]) return 1;
    return 0;
}

int dfs(int u, int flow)
{
    if (u == T || !flow) return flow;
    int d, used = 0;
    for (int i = cur[u]; ~i; i = edge[i].nxt)
    {
        cur[u] = i;
        int v = edge[i].to;
        if (dep[v] == dep[u] + 1 && (d = dfs(v, min(flow, edge[i].dis))))
        {
            used += d;
            flow -= d;
            edge[i].dis -= d;
            edge[i ^ 1].dis += d;
            if (!flow) break;
        }
    }
    if (!used) dep[u] = -2;
    return used;
}

int dinic()
{
    int ans = 0;
    while (bfs(S, T)) ans += dfs(S, inf);
    return ans;
}

void write(int x)
{
    if (x < 0)
    {
        putchar(‘-‘);
        x = -x;
    }
    if (x > 9) write(x / 10);
    putchar(x % 10 + ‘0‘);
}

int main()
{
    origin();
    n = read(), p = read(), q = read();
    S = 0, T = 500;
    rep(i, 1, p)
    {
        addedge(S, i, 1);
        addedge(i, S, 0);
    }
    rep(i, 1, q)
    {
        addedge(i + p, T, 1);
        addedge(T, i + p, 0);
    }
    rep(i, 1, n)
    {
        addedge(i + p + q, i + p + q + n, 1);
        addedge(i + p + q + n, i + p + q, 0);
    }
    rep(i, 1, n)
    {
        rep(j, 1, p)
        {
            int x;
            x = read();
            if (x == 1)
            {
                addedge(j, i + p + q, 1);
                addedge(i + p + q, j, 0);
            }
        }
    }
    rep(i, 1, n)
    {
        rep(j, 1, q)
        {
            int x;
            x = read();
            if (x == 1)
            {
                addedge(i + p + q + n, j + p, 1);
                addedge(j + p, i + p + q + n, 0);
            }
        }
    }
    ans = dinic();
    write(ans);
    return 0;
}

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