[luoguP2765] 魔术球问题（最大流—最小不相交路径覆盖）

传送门

枚举球的个数 num

如果 i < j && (i + j) 是完全平方数，那么 i -> j' 连一条边

再加一个超级源点 s，s -> i

再加一个超级汇点 t，i' -> t

那么当前可以放的柱子的最小数量就是最小不相交路径数

如果当前的最小不相交路径数 > num，break

求最大流的时候别忘了记录方案

——代码

 #include <cmath>
 #include <queue>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #define N 10001
 #define M 200001
 #define mid 5000
 #define min(x, y) ((x) < (y) ? (x) : (y))

 int n, cnt, sum, ans, s, t = mid << ;
 int head[N], to[M], next[M], val[M], suc[N], dis[N];
 bool vis[N];

 inline int read()
 {
     int x = , f = ;
     char ch = getchar();
     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -;
     for(; isdigit(ch); ch = getchar()) x = (x << ) + (x << ) + ch - '';
     return x * f;
 }

 inline void add(int x, int y, int z)
 {
     to[cnt] = y;
     val[cnt] = z;
     next[cnt] = head[x];
     head[x] = cnt++;
 }

 inline bool bfs()
 {
     int i, u, v;
     std::queue <int> q;
     memset(dis, -, sizeof(dis));
     q.push(s);
     dis[s] = ;
     while(!q.empty())
     {
         u = q.front(), q.pop();
         for(i = head[u]; i ^ -; i = next[i])
         {
             v = to[i];
             if(val[i] && dis[v] == -)
             {
                 dis[v] = dis[u] + ;
                 if(v == t) return ;
                 q.push(v);
             }
         }
     }
     return ;
 }

 inline int dfs(int u, int maxflow)
 {
     if(u == t) return maxflow;
     int i, v, d, ret = ;
     for(i = head[u]; i ^ -; i = next[i])
     {
         v = to[i];
         if(val[i] && dis[v] == dis[u] + )
         {
             d = dfs(v, min(val[i], maxflow - ret));
             ret += d;
             val[i] -= d;
             val[i ^ ] += d;
             if(d) suc[u] = v - mid;
             if(ret == maxflow) return ret;
         }
     }
     return ret;
 }

 int main()
 {
     int i, j, now, num = ;
     n = read();
     memset(head, -, sizeof(head));
     while()
     {
         num++;
         add(s, num, ), add(num, s, );
         add(num + mid, t, ), add(t, num + mid, );
         for(i = ; i < num; i++)
             if(sqrt(i + num) == (int)sqrt(i + num))
                 add(i, num + mid, ), add(num + mid, i, );
         while(bfs()) sum += dfs(s, 1e9);
         if(num - sum > n) break;
     }
     cnt = ;
     memset(head, -, sizeof(head));
     for(i = ; i < num; i++)
     {
         add(s, i, ), add(i, s, );
         add(i + mid, t, ), add(t, i + mid, );
     }
     for(i = ; i < num; i++)
         for(j = ; j < i; j++)
             if(sqrt(i + j) == (int)sqrt(i + j))
                 add(j, i + mid, ), add(i + mid, j, );
     while(bfs()) dfs(s, 1e9);
     printf("%d\n", num - );
     for(i = ; i < num; i++)
         if(!vis[i])
         {
             now = i;
             while(now)
             {
                 printf("%d ", now);
                 vis[now] = ;
                 now = suc[now];
             }
             puts("");
         }
     return ;
 }