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。

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cmath>
  4 #include<algorithm>
  5 #include<cstring>
  6 #include<cstdlib>
  7 #include<cctype>
  8 #include<vector>
  9 #include<stack>
 10 #include<queue>
 11 using namespace std;
 12 #define enter puts("") 
 13 #define space putchar(' ')
 14 #define Mem(a) memset(a, 0, sizeof(a))
 15 typedef long long ll;
 16 typedef double db;
 17 const int INF = 0x3f3f3f3f;
 18 const db eps = 1e-8;
 19 const int maxn = 455;
 20 inline ll read()
 21 {
 22     ll ans = 0;
 23     char ch = getchar(), last = ' ';
 24     while(!isdigit(ch)) {last = ch; ch = getchar();}
 25     while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
 26     if(last == '-') ans = -ans;
 27     return ans;
 28 }
 29 inline void write(ll x)
 30 {
 31     if(x < 0) x = -x, putchar('-');
 32     if(x >= 10) write(x / 10);
 33     putchar(x % 10 + '0');
 34 }
 35 
 36 int m, n, t, Sum;
 37 struct Edge
 38 {
 39     int from, to, cap, flow;
 40 };
 41 vector<Edge> edges;
 42 vector<int> G[maxn];
 43 void addEdge(int from, int to, int w)
 44 {
 45     edges.push_back((Edge){from, to, w, 0});
 46     edges.push_back((Edge){to, from, 0, 0});
 47     int sz = edges.size();
 48     G[from].push_back(sz - 2);
 49     G[to].push_back(sz - 1);
 50 }
 51 
 52 int dis[maxn];
 53 bool bfs()
 54 {
 55     Mem(dis); dis[0] = 1;
 56     queue<int> q; q.push(0);
 57     while(!q.empty())
 58     {
 59         int now = q.front(); q.pop();
 60         for(int i = 0; i < (int)G[now].size(); ++i)
 61         {
 62             Edge& e = edges[G[now][i]];
 63             if(!dis[e.to] && e.cap > e.flow)
 64             {
 65                 dis[e.to] = dis[now] + 1;
 66                 q.push(e.to);
 67             }
 68         }
 69     }
 70     return dis[t];
 71 }
 72 int cur[maxn];
 73 int dfs(int now, int res)
 74 {
 75     if(now == t || res == 0) return res;
 76     int flow = 0, f;
 77     for(int& i = cur[now]; i < (int)G[now].size(); ++i)
 78     {
 79         Edge& e = edges[G[now][i]];
 80         if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(res, e.cap - e.flow))) > 0)
 81         {
 82             e.flow += f;
 83             edges[G[now][i] ^ 1].flow -= f;
 84             flow += f;
 85             res -= f;
 86             if(res == 0) break;
 87         }
 88     }
 89     return flow;
 90 }
 91 
 92 int maxflow()
 93 {
 94     int flow = 0;
 95     while(bfs())
 96     {
 97         Mem(cur);
 98         flow += dfs(0, INF);
 99     }
100     return flow;
101 }
102 
103 int main()
104 {
105     m = read(); n = read();
106     t = n + m + 1;
107     for(int i = 1; i <= m; ++i)
108     {
109         int x = read(); Sum += x;
110         addEdge(0, i, x);
111         for(int j = 1; j <= n; ++j) addEdge(i, j + m, 1);
112     }
113     for(int i = 1; i <= n; ++i) addEdge(i + m, t, read());
114     if(maxflow() == Sum) 
115     {
116         write(1); enter;
117         for(int i = 1; i <= m; ++i)
118         {
119             for(int j = 0; j < (int)G[i].size(); ++j)
120             {
121                 Edge e = edges[G[i][j]];
122                 if(e.to > m && e.cap - e.flow == 0)
123                     write(e.to - m), space;
124             }
125             enter;
126         }
127     }
128     else write(0), enter;
129     return 0;
130 }

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