POJ 1273 最大流

POJ 1273 最大流

一、题目描述

Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.

Input

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

Sample Input

5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

Sample Output

50

二、分析
   其实就是以点1为s，以点m为t，求最大流，但是要注意输入的路径可以重复(见代码30行)，使用Edmonds-Karp算法，具体算法： 最大流问题
三、代码

 1 #include < iostream >
 2 #include < queue >
 3 using   namespace  std;
 4 #define  MAXM 201
 5 int  m, n;
 6 int  si, ei, ci;
 7 int  c[MAXM][MAXM];
 8 int  f[MAXM][MAXM];
 9 int  cf[MAXM][MAXM];
10 bool  visit[MAXM];
11 int  p[MAXM];
12 struct  node
13 {
14    int v, cf;
15    void set(int vv, int ccf)
16    {
17        v = vv; cf = ccf;
18    }
19} ;
20 int  main()
21 {
22    while(scanf("%d%d", &n, &m) != EOF)
23    {
24        memset(c, 0, sizeof(c));
25        memset(f, 0, sizeof(f));
26        memset(cf, 0, sizeof(cf));
27        while(n--)
28        {
29            scanf("%d%d%d", &si, &ei, &ci);
30            c[si][ei] += ci;
31            cf[si][ei] = c[si][ei];
32        }
33        bool flag = true; //用于表示是否找到增广路
34        while(flag)
35        {
36            flag = false;
37            memset(visit, 0, sizeof(visit));
38            queue<node> q;
39            node temp;
40            temp.set(1, INT_MAX);
41            p[1] = 0;
42            q.push(temp); visit[1] = true;
43            while(!q.empty()) //广度优先搜索
44            {
45                node temp = q.front(); q.pop();
46                for(int i=1; i<=m; i++)
47                {
48                    if(temp.v == i || visit[i] || cf[temp.v][i] == 0)
49                        continue;
50                    node newNode; 
51                    newNode.set(i, min(temp.cf, cf[temp.v][i]));
52                    p[i] = temp.v;
53                    q.push(newNode);
54                    visit[i] = true;
55                    if(i == m)
56                    {
57                        flag = true; //找到增广路
58                        break;
59                    }
60                }
61                if(flag)
62                    break;
63            }
64            if(flag)
65            {
66                int mincf = q.back().cf;
67                int v1 = p[m], v2 = m;
68                while(v1 != 0)
69                {
70                    f[v1][v2] += mincf; //修改流
71                    f[v2][v1] = -f[v1][v2];
72                    cf[v1][v2] = c[v1][v2] - f[v1][v2]; //修改残留容量
73                    cf[v2][v1] = c[v2][v1] - f[v2][v1];
74                    v2 = v1;
75                    v1 = p[v1];
76                }
77            }
78        }
79        int res = 0;
80        for(int i=2; i<=m; i++) //计算最大流
81            res += f[1][i];
82        printf("%d\n", res);
83    }
84}