網路流24題

洛谷題單：網路流24題

第一題：飛行員配對問題

二分圖模板題，匈牙利可以過，並且匈牙利演算法可以記錄匹配的資訊

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <queue>
 5 #define N 100010
 6 using namespace std;
 7 int cnt, n, m, head[N], nxt[N], to[N], p[N];
 8 bool vis[N];
 9 inline int read() {
 
10     int x = 0, y = 1;
11     char c = getchar();
12     while (c < '0' || c > '9') {
13         if (c == '-')
14             y = -1;
15         c = getchar();
16     }
17     while (c >= '0' && c <= '9')
18         x = x * 10 + c - '0', c = getchar();
19     return x * y;
20 }
21 void
 
 add(int x, int y) {
22     nxt[++cnt] = head[x];
23     head[x] = cnt;
24     to[cnt] = y;
25 }
26 bool match(int u) {
27     for (int i = head[u]; ~i; i = nxt[i]) {
28         int v = to[i];
29         if (vis[v])
30             continue;
31         else
32             vis[v] = true;
33         if (p[v] == 0
 
 || match(p[v])) {
34             p[v] = u;
35             return true;
36         }
37     }
38     return false;
39 }
40 int pipei() {
41     int ans = 0;
42     for (int i = 1; i <= m; i++) {
43         memset(vis, false, sizeof(vis));
44         if (match(i))
45             ans++;
46     }
47     return ans;
48 }
49 int main() {
50     cnt = -1;
51     memset(head, -1, sizeof(head)); memset(p,0,sizeof(p));
52     m = read();
53     n = read();
54     int u, v;
55     while (~scanf("%d%d", &u, &v) && u+v != -2) add(u, v);
56     int k = pipei();
57     printf("%d\n",k);
58     for(int i = m+1; i <= n; i++)
59         if(p[i] != 0)
60             printf("%d %d\n",p[i],i);
61     return 0;
62 }

第二題：太空飛行計劃問題

最小割的應用，最大權閉合子圖模板題，源點建正權，匯點建負權

最大權閉合子圖的權值和=max{被選擇的點權和}=正點權和−min{沒被選擇的正權點之和+被選擇的負權點絕對值和}

=正點權和−最小割

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <queue>
  5 #include <vector>
  6 #define N 10010
  7 using namespace std;
  8 vector<int> g[N];
  9 int cnt,n,m,s,t,head[N],nxt[N],to[N],d[N],cur[N],w[N],f[N];
 10 const int INF = 0x3f3f3f3f;
 11 void add(int x, int y, int z)
 12 {
 13     nxt[++cnt] = head[x];
 14     head[x] = cnt;
 15     to[cnt] = y;
 16     w[cnt] = z;
 17 }
 18 bool bfs(int s, int t)
 19 {
 20     memset(d,-1,sizeof(d));
 21     d[s] = 0;
 22     queue<int> q;
 23     q.push(s);
 24     while(!q.empty())
 25     {
 26         int p = q.front(); q.pop();
 27         for(int i = head[p]; ~i; i = nxt[i])
 28         {
 29             int v = to[i];
 30             if(d[v] == -1 && w[i])
 31             {
 32                 d[v] = d[p]+1;
 33                 q.push(v);
 34             }
 35         }
 36     }
 37     return d[t] != -1;
 38 }
 39 int dfs(int s, int flow)
 40 {
 41     if(s == t) return flow;
 42     int ans = 0;
 43     for(int i = cur[s]; ~i; i = nxt[i])
 44     {
 45         int v = to[i];
 46         cur[s] = i;
 47         if(d[v] == d[s]+1 && w[i])
 48         {
 49             int dis = dfs(v,min(flow,w[i]));
 50             if(dis)
 51             {
 52                 w[i] -= dis;
 53                 w[i^1] += dis;
 54                 flow -= dis;
 55                 ans += dis;
 56                 if(flow == 0) break;
 57             }
 58         }
 59     }
 60     return ans;
 61 }
 62 int dinic(int s, int t)
 63 {
 64     int ans = 0;
 65     while(bfs(s,t))
 66     {
 67         memcpy(cur,head,sizeof(head));
 68         ans += dfs(s,INF);
 69     }
 70     return ans;
 71 }
 72 int main()
 73 {
 74     int sum = 0;
 75     cnt = -1; memset(head,-1,sizeof(head));
 76     scanf("%d%d",&m,&n);
 77     s = 0; t = n+m+1;
 78     for(int i = 1; i <= m; i++)
 79     {
 80         int x; char c;
 81         scanf("%d",&x);
 82         sum += x;
 83         add(s,i+n,x); add(i+n,s,0);
 84         while(true)
 85         {
 86             scanf("%d%c",&x,&c);
 87             add(i+n,x,INF); add(x,i+n,0);
 88             if(c == '\n' || c == '\r') break;
 89         }
 90     }
 91     for(int i = 1; i <= n; i++)
 92     {
 93         int x;
 94         scanf("%d",&x);
 95         add(i,t,x); add(t,i,0);
 96     }
 97     sum -= dinic(s,t);
 98     for(int i = 1; i <= m; i++)
 99         if(d[i+n] != -1)
100             printf("%d ",i);
101     puts("");
102     for(int i = 1; i <= n; i++)
103         if(d[i] != -1)
104             printf("%d ",i);
105     puts("");
106     printf("%d\n",sum);
107     return 0;
108 }