圆桌问题

嘟嘟嘟

这道题建图还是很明白的：一个二分图，左边是单位，右边是餐桌，然后源点都向单位连一条容量为单位人数的边，每一个餐桌和汇点都连一条容量为餐桌容量的边（都叫容量~）。然后每一个单位向每一个餐桌都连一条容量为1的边，跑最大流即可。本来以为复杂度会很高，然而数据范围小，加上dinic玄学，很快就跑过了。

最后输出答案：如果最大流等于所有单位人数的话就说明满足要求，然后查询每一个单位的边的流量情况：如果这条边流满了，就说明该单位有一个人到了这一桌，那么输出该桌的编号。

后来知道这叫二分图多重匹配。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a) memset(a, 0, sizeof(a))
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 455;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

int m, n, t, Sum;
struct Edge
{
    int from, to, cap, flow;
};
vector<Edge> edges;
vector<int> G[maxn];
void addEdge(int from, int to, int w)
{
    edges.push_back((Edge){from, to, w, 0});
    edges.push_back((Edge){to, from, 0, 0});
    int sz = edges.size();
    G[from].push_back(sz - 2);
    G[to].push_back(sz - 1);
}

int dis[maxn];
bool bfs()
{
    Mem(dis); dis[0] = 1;
    queue<int> q; q.push(0);
    while(!q.empty())
    {
        int now = q.front(); q.pop();
        for(int i = 0; i < (int)G[now].size(); ++i)
        {
            Edge& e = edges[G[now][i]];
            if(!dis[e.to] && e.cap > e.flow)
            {
                dis[e.to] = dis[now] + 1;
                q.push(e.to);
            }
        }
    }
    return dis[t];
}
int cur[maxn];
int dfs(int now, int res)
{
    if(now == t || res == 0) return res;
    int flow = 0, f;
    for(int& i = cur[now]; i < (int)G[now].size(); ++i)
    {
        Edge& e = edges[G[now][i]];
        if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(res, e.cap - e.flow))) > 0)
        {
            e.flow += f;
            edges[G[now][i] ^ 1].flow -= f;
            flow += f;
            res -= f;
            if(res == 0) break;
        }
    }
    return flow;
}

int maxflow()
{
    int flow = 0;
    while(bfs())
    {
        Mem(cur);
        flow += dfs(0, INF);
    }
    return flow;
}

int main()
{
    m = read(); n = read();
    t = n + m + 1;
    for(int i = 1; i <= m; ++i)
    {
        int x = read(); Sum += x;
        addEdge(0, i, x);
        for(int j = 1; j <= n; ++j) addEdge(i, j + m, 1);
    }
    for(int i = 1; i <= n; ++i) addEdge(i + m, t, read());
    if(maxflow() == Sum) 
    {
        write(1); enter;
        for(int i = 1; i <= m; ++i)
        {
            for(int j = 0; j < (int)G[i].size(); ++j)
            {
                Edge e = edges[G[i][j]];
                if(e.to > m && e.cap - e.flow == 0)
                    write(e.to - m), space;
            }
            enter;
        }
    }
    else write(0), enter;
    return 0;
}