网络流模板

知识点:https://blog.csdn.net/x_y_q_/article/details/51999466

https://www.cnblogs.com/ZJUT-jiangnan/p/3632525.html

题集:https://blog.csdn.net/shahdza/article/details/7779537

 

1532.最大流

语法:https://blog.csdn.net/xiaofei0859/article/details/78452510

EK算法

 1 #include <iostream>
 2 #include <queue>
 3 #include<string.h>
 4 using namespace std;
 5 #define arraysize 201
 6 int maxData = 0x7fffffff;
 7 int capacity[arraysize][arraysize]; //记录残留网络的容量
 8 int flow[arraysize];                //标记从源点到当前节点实际还剩多少流量可用
 9 int pre[arraysize];                 //标记在这条路径上当前节点的前驱,同时标记该节点是否在队列中
10 int n,m;//m是汇点,n是边数
11 queue<int> myqueue;
12 int BFS(int src,int des)
13 {
14     int i,j;
15     while(!myqueue.empty())       //队列清空
16         myqueue.pop();
17     for(i=1;i<m+1;++i)
18     {
19         pre[i]=-1;
20     }
21     pre[src]=0;
22     flow[src]= maxData;
23     myqueue.push(src);
24     while(!myqueue.empty())
25     {
26         int index = myqueue.front();
27         myqueue.pop();
28         if(index == des)            //找到了增广路径
29             break;
30         for(i=1;i<m+1;++i)
31         {
32             if(i!=src && capacity[index][i]>0 && pre[i]==-1)
33             {
34                  pre[i] = index; //记录前驱
35                  flow[i] = min(capacity[index][i],flow[index]);   //关键:迭代的找到增量
36                  myqueue.push(i);
37             }
38         }
39     }
40     if(pre[des]==-1)      //残留图中不再存在增广路径
41         return -1;
42     else
43         return flow[des];//该流大小
44 }
45 int maxFlow(int src,int des)
46 {
47     int increasement= 0;
48     int sumflow = 0;
49     while((increasement=BFS(src,des))!=-1)
50     {
51          int k = des;          //利用前驱寻找路径
52          while(k!=src)
53          {
54               int last = pre[k];
55               capacity[last][k] -= increasement; //改变正向边的容量
56               capacity[k][last] += increasement; //改变反向边的容量
57               k = last;
58          }
59          sumflow += increasement;
60     }
61     return sumflow;
62 }
63 int main()
64 {
65     int i,j;
66     int start,end,ci;
67     while(cin>>n>>m)
68     {
69         memset(capacity,0,sizeof(capacity));
70         memset(flow,0,sizeof(flow));
71         for(i=0;i<n;++i)
72         {
73             cin>>start>>end>>ci;
74             if(start == end)               //考虑起点终点相同的情况
75                continue;
76             capacity[start][end] +=ci;     //此处注意可能出现多条同一起点终点的情况
77         }
78         cout<<maxFlow(1,m)<<endl;
79     }
80     return 0;
81 }

 

Dinic算法

知识点:http://www.cnblogs.com/SYCstudio/p/7260613.html

https://www.cnblogs.com/LUO77/p/6115057.html

https://www.cnblogs.com/y-clever/p/6308820.html

https://blog.csdn.net/baidu_35009437/article/details/52068468

https://comzyh.com/blog/archives/568/

 1 #include <cstdio>
 2 #include <string.h>
 3 #include <queue>
 4 using namespace std;
 5 int const inf = 0x3f3f3f3f;
 6 int const MAX = 205;
 7 int n, m;//m是汇点,n是边数
 8 int c[MAX][MAX], dep[MAX];//dep[MAX]代表当前层数
 9 
10 int bfs(int s, int t)//重新建图,按层次建图
11 {
12     queue<int> q;
13     while(!q.empty())
14         q.pop();
15     memset(dep, -1, sizeof(dep));
16     dep[s] = 0;
17     q.push(s);
18     while(!q.empty()){
19         int u = q.front();
20         q.pop();
21         for(int v = 1; v <= m; v++){
22             if(c[u][v] > 0 && dep[v] == -1){//如果可以到达且还没有访问,可以到达的条件是剩余容量大于0,没有访问的条件是当前层数还未知
23                 dep[v] = dep[u] + 1;
24                 q.push(v);
25             }
26         }
27     }
28     return dep[t] != -1;//如等于-1则为无法分层的图
29 }
30 
31 int dfs(int u, int mi, int t)//查找路径上的最小流量
32 {
33     if(u == t)
34         return mi;
35     int tmp;
36     for(int v = 1; v <= m; v++){
37         if(c[u][v] > 0 && dep[v] == dep[u] + 1  && (tmp = dfs(v, min(mi, c[u][v]), t))){
38             c[u][v] -= tmp;
39             c[v][u] += tmp;
40             return tmp;
41         }
42     }
43     return 0;
44 }
45 
46 int dinic()
47 {
48     int ans = 0, tmp;
49     while(bfs(1, m)){
50         while(1){
51             tmp = dfs(1, inf, m);
52             if(tmp == 0)
53                 break;
54             ans += tmp;
55         }
56     }
57     return ans;
58 }
59 
60 int main()
61 {
62     while(~scanf("%d %d", &n, &m)){
63         memset(c, 0, sizeof(c));
64         int u, v, w;
65         while(n--){
66             scanf("%d %d %d", &u, &v, &w);
67             c[u][v] += w;
68         }
69         printf("%d\n", dinic());
70     }
71     return 0;
72 }

 

posted @ 2018-08-22 20:49  hemeiwolong  阅读(142)  评论(0)    收藏  举报