[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

  1 #include <cmath>
  2 #include <queue>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <iostream>
  6 #define N 10001
  7 #define M 200001
  8 #define mid 5000
  9 #define min(x, y) ((x) < (y) ? (x) : (y))
 10 
 11 int n, cnt, sum, ans, s, t = mid << 1;
 12 int head[N], to[M], next[M], val[M], suc[N], dis[N];
 13 bool vis[N];
 14 
 15 inline int read()
 16 {
 17     int x = 0, f = 1;
 18     char ch = getchar();
 19     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
 20     for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
 21     return x * f;
 22 }
 23 
 24 inline void add(int x, int y, int z)
 25 {
 26     to[cnt] = y;
 27     val[cnt] = z;
 28     next[cnt] = head[x];
 29     head[x] = cnt++;
 30 }
 31 
 32 inline bool bfs()
 33 {
 34     int i, u, v;
 35     std::queue <int> q;
 36     memset(dis, -1, sizeof(dis));
 37     q.push(s);
 38     dis[s] = 0;
 39     while(!q.empty())
 40     {
 41         u = q.front(), q.pop();
 42         for(i = head[u]; i ^ -1; i = next[i])
 43         {
 44             v = to[i];
 45             if(val[i] && dis[v] == -1)
 46             {
 47                 dis[v] = dis[u] + 1;
 48                 if(v == t) return 1;
 49                 q.push(v);
 50             }
 51         }
 52     }
 53     return 0;
 54 }
 55 
 56 inline int dfs(int u, int maxflow)
 57 {
 58     if(u == t) return maxflow;
 59     int i, v, d, ret = 0;
 60     for(i = head[u]; i ^ -1; i = next[i])
 61     {
 62         v = to[i];
 63         if(val[i] && dis[v] == dis[u] + 1)
 64         {
 65             d = dfs(v, min(val[i], maxflow - ret));
 66             ret += d;
 67             val[i] -= d;
 68             val[i ^ 1] += d;
 69             if(d) suc[u] = v - mid;
 70             if(ret == maxflow) return ret;
 71         }
 72     }
 73     return ret;
 74 }
 75 
 76 int main()
 77 {
 78     int i, j, now, num = 0;
 79     n = read();
 80     memset(head, -1, sizeof(head));
 81     while(1)
 82     {
 83         num++;
 84         add(s, num, 1), add(num, s, 0);
 85         add(num + mid, t, 1), add(t, num + mid, 0);
 86         for(i = 1; i < num; i++)
 87             if(sqrt(i + num) == (int)sqrt(i + num))
 88                 add(i, num + mid, 1), add(num + mid, i, 0);
 89         while(bfs()) sum += dfs(s, 1e9);
 90         if(num - sum > n) break;
 91     }
 92     cnt = 0;
 93     memset(head, -1, sizeof(head));
 94     for(i = 1; i < num; i++)
 95     {
 96         add(s, i, 1), add(i, s, 0);
 97         add(i + mid, t, 1), add(t, i + mid, 0);
 98     }
 99     for(i = 1; i < num; i++)
100         for(j = 1; j < i; j++)
101             if(sqrt(i + j) == (int)sqrt(i + j))
102                 add(j, i + mid, 1), add(i + mid, j, 0);
103     while(bfs()) dfs(s, 1e9);
104     printf("%d\n", num - 1);
105     for(i = 1; i < num; i++)
106         if(!vis[i])
107         {
108             now = i;
109             while(now)
110             {
111                 printf("%d ", now);
112                 vis[now] = 1;
113                 now = suc[now];
114             }
115             puts("");
116         }
117     return 0;
118 }
View Code

 

posted @ 2017-06-02 09:58  zht467  阅读(217)  评论(0)  编辑  收藏  举报