Flow Problem
Time Limit: 5000/5000 MS
(Java/Others) Memory Limit: 65535/32768 K
(Java/Others)
Total Submission(s): 6817 Accepted
Submission(s): 3178
Problem Description
Network flow is a well-known difficult problem for
ACMers. Given a graph, your task is to find out the maximum flow for the
weighted directed graph.
Input
The first line of input contains an integer T,
denoting the number of test cases.
For each test case, the first line
contains two integers N and M, denoting the number of vertexes and edges in the
graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line
contains three integers X, Y and C, there is an edge from X to Y and the
capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
Output
For each test cases, you should output the maximum
flow from source 1 to sink N.
Sample Input
2 3 2 1 2 1 2 3 1 3 3 1 2 1 2 3 1 1 3 1
Sample Output
Case 1: 1 Case 2: 2
Author
HyperHexagon
Source
HyperHexagon‘s
Summer Gift (Original tasks)
题意：最大流
分析：最大流模板
代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <queue>
5 #include <vector>
6 using namespace std;
7
8 const int maxn = 20 << 1;
9 const int INF = 1000000000;
10
11 struct Edge
12 {
13 int from, to, cap, flow;
14 };
15
16 struct Dinic
17 {
18 int n, m, s, t;
19 vector<Edge> edges;
20 vector<int>G[maxn];
21 bool vis[maxn];
22 int d[maxn];
23 int cur[maxn];
24
25 void ClearAll(int n) {
26 for(int i = 0; i <= n; i++) {
27 G[i].clear();
28 }
29 edges.clear();
30 }
31
32 void AddEdge(int from, int to, int cap) {
33 edges.push_back((Edge){from, to, cap, 0} );
34 edges.push_back((Edge){to, from, 0, 0} );
35 m = edges.size();
36 G[from].push_back(m - 2);
37 G[to].push_back(m - 1);
38 //printf("%din end\n",m);
39 }
40
41 bool BFS()
42 {
43 memset(vis, 0, sizeof(vis) );
44 queue<int> Q;
45 Q.push(s);
46 vis[s] = 1;
47 d[s] = 0;
48 while(!Q.empty() ){
49 int x = Q.front(); Q.pop();
50 for(int i = 0; i < G[x].size(); i++) {
51 Edge& e = edges[G[x][i]];
52 if(!vis[e.to] && e.cap > e.flow) {
53 vis[e.to] = 1;
54 d[e.to] = d[x] + 1;
55 Q.push(e.to);
56 }
57 }
58 }
59 return vis[t];
60 }
61
62 int DFS(int x, int a) {
63 if(x == t || a == 0) return a;
64 int flow = 0, f;
65 for(int& i = cur[x]; i < G[x].size(); i++) {
66 Edge& e = edges[G[x][i]];
67 if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
68 e.flow += f;
69 edges[G[x][i]^1].flow -= f;
70 flow += f;
71 a -= f;
72 if(a == 0) break;
73 }
74 }
75 return flow;
76 }
77
78 int Maxflow(int s, int t) {
79 this -> s = s; this -> t = t;
80 int flow = 0;
81 while(BFS()) {
82 memset(cur, 0, sizeof(cur) );
83 flow += DFS(s, INF);
84 }
85 return flow;
86 }
87 };
88
89 Dinic g;
90
91 int main()
92 {
93 int t;
94 scanf("%d",&t);
95 int n, m;
96 int a, b, c;
97 for(int kase = 1; kase <= t; kase++) {
98 scanf("%d %d",&n, &m);
99 g.ClearAll(maxn);
100 for(int i = 0;i < m; i++ ) {
101 scanf("%d %d %d",&a, &b, &c);
102 g.AddEdge(a, b, c);
103 }
104 printf("Case %d: %d\n",kase, g.Maxflow(1, n) );
105 }
106 return 0;
107 }
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Code
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