Flow Problem
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1252 Accepted Submission(s): 606
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
首先给定有多少组测试数据，然后给定一个顶点数，边数，然后给定一个单向的流量，求从第一个点到最后一个点的最大流量。注意该流网络是有向图，所以对于流量的控制是给定x 到 y 的流量是 z ， 那么只需要给cap[x][y]+= z 即可。求解网络流就是在图中寻找增广路径，一条增广路径是指能够从起点到终点寻找到一条大于零的流量路径，并试图将其填充满，注意抑制一次寻找过程中回流现象，这是无意义的，因为在每一次队列的计算中，并没有及时的更新管道的容量，所以回流的形成会无限制的循环，肯定会超出内存，实质上也是一个死循环。
代码如下：
#include#include #include #include #include #define INF 0x7f7f7f7f usingnamespace std; int flow[205][205], cap[205][205], p[205], c[205]; // f用来记录单前节点流，p用来记录父路径 void BFS( int&maxflow, int N ) { queue q; maxflow=0; bool finish=false; memset( flow, 0, sizeof( flow ) ); while( !finish ) { memset( p, 0, sizeof( p ) ); memset( c, 0, sizeof( c ) ); c[1]= INF, p[1]=-1; q.push( 1 ); while( !q.empty() ) { int pos= q.front(); q.pop(); for( int i=1; i<= N; ++i ) { if( !c[i]&& flow[pos][i]< cap[pos][i] ) // 如果该条边没有达到饱和或者形成回流 { c[i]= min( c[pos], cap[pos][i]- flow[pos][i] ); // 选取当前管道残留容量和截止到上一次的路径残留容量之间的较小者，可以看作是一种更新 // 即该增广路径的残留容量是有路径中残留容量最小的管道决定的 p[i]= pos; // 记录父亲路径 q.push( i ); } } } if( c[N]==0 ) { finish=true; } maxflow+= c[N]; int pos= N; while( !finish&& pos!=1 ) { flow[ p[pos] ][ pos ]+= c[N]; flow[ pos ][ p[pos] ]-= c[N]; pos= p[ pos ]; } } } int main() { int N, M, maxflow, ca=0, T; scanf( "%d", &T ); while( T-- ) { scanf( "%d %d", &M, &N ); memset( cap, 0, sizeof( cap ) ); for( int i=1; i<= N; ++i ) { int x, y, z; scanf( "%d %d %d", &x, &y, &z ); cap[x][y]+= z; } BFS( maxflow, M ); printf( "Case %d: %d\n", ++ca, maxflow ); } }
Dinic版 Dinic算法相比于上面的KM算法，采用了分层思想，以最短路寻找增广路，时间复杂度O(V2E)。
#include#include #include #include #define INF 0x3fffffff#define RE(x) (x)^1#define S 1#define MAXN 20using namespace std;int N, M, head[MAXN], lvl[MAXN], idx;struct Edge{ int No, cap, next;}e[1005];void init(){ idx = -1; memset(head, 0xff, sizeof (head));}void insert(int x, int y, int z){ ++idx; e[idx].No = y; e[idx].cap = z; e[idx].next = head[x]; head[x] = idx; ++idx; e[idx].No = x; e[idx].cap = 0; e[idx].next = head[y]; head[y] = idx;}bool bfs(){ int pos; queue q; memset(lvl, 0xff, sizeof (lvl)); lvl[S] = 0; q.push(S); while (!q.empty()) { pos = q.front(); q.pop(); for (int i = head[pos]; i!=-1; i = e[i].next) { if (e[i].cap > 0 && lvl[e[i].No]==-1) { lvl[e[i].No] = lvl[pos]+1; q.push(e[i].No); // 注意不要直接把“i”push进去了 } } } return lvl[N] != -1;}int dfs(int u, int flow){ if (u == N) { return flow; // 结束条件，把流推到了T节点（汇点） } int tf = 0, f; for (int i = head[u]; i != -1; i = e[i].next) { if (lvl[u]+1 == lvl[e[i].No] && e[i].cap > 0 && (f = dfs(e[i].No, min(e[i].cap, flow - tf)))) { e[i].cap -= f; e[RE(i)].cap += f; tf += f; } } if (tf == 0) { lvl[u] = -1; // tf等于零表示该点没有进行增广的能力，应及时将其高度赋值为-1，防止第二次被搜索到 } return tf;}int main(){ int T, ans, ca = 0, a, b, c; scanf("%d", &T); while (T--) { init(); ans = 0; scanf("%d %d", &N, &M); for (int i = 0; i < M; ++i) { scanf("%d %d %d", &a, &b, &c); insert(a, b, c); } while (bfs()) { ans += dfs(S, INF); } printf("Case %d: %d\n", ++ca, ans); } return 0; }
寻找增广路径的sap算法，时间复杂度为O(V2E)：
#include#include #include #include #include #define INF 0x3fffffff#define RE(x) (x)^1#define MAXN 20using namespace std;int lvl[MAXN], gap[MAXN], idx, head[MAXN], sink, source, N, M;// lvl用来记录每个节点离汇点的高度，gap数组用来记录高度值为i的个数gap[i]struct Edge{ int v, cap, next;}e[2010];void init(){ idx = -1; memset(head, 0xff, sizeof (head)); memset(lvl, 0, sizeof (lvl)); memset(gap, 0, sizeof (gap)); // 0xff == -1 (对于一个字节来说)}void insert(int a, int b, int c){ ++idx; e[idx].v = b, e[idx].next = head[a]; e[idx].cap = c, head[a] = idx;}int dfs(int u, int flow){ if (u == sink) { return flow; } int tf = 0, sf, mlvl = N-1; for (int i = head[u]; i != -1; i = e[i].next) { if (e[i].cap > 0) { // 只要还有边还有容量的话 if (lvl[u] == lvl[e[i].v]+1) { // 与dinic不同这个层数是反的 sf = dfs(e[i].v, min(flow-tf, e[i].cap)); e[i].cap -= sf; e[RE(i)].cap += sf; tf += sf; if (lvl[source] >= N) { // 可能在子递归中出现了断层或者不能再流的现象，这个时候要及时的退出 return tf; } if (tf == flow) { break; } } mlvl = min(mlvl, lvl[e[i].v]); } } if (tf == 0) { --gap[lvl[u]]; if (!gap[lvl[u]]) { lvl[source] = N; } else { lvl[u] = mlvl+1; ++gap[lvl[u]]; } } return tf;}int sap(){ int ans = 0; gap[0] = N; while (lvl[source] < N) { ans += dfs(source, INF); } return ans;}int main(){ int T, a, b, c, ca = 0, ans; scanf("%d", &T); while (T--) { init(); ans = 0; scanf("%d %d", &N, &M); for (int i = 0; i < M; ++i) { scanf("%d %d %d", &a, &b, &c); insert(a, b, c); insert(b, a, 0); } sink = N, source = 1; printf("Case %d: %d\n", ++ca, sap()); } return 0;}
网络流 sap非递归模板：
// HDU-4280 #include#include #include #define VM 100010#define EM 400010const int inf = 0x3f3f3f3f;struct E{ int to, frm, nxt, cap;}edge[EM];int head[VM],e,n,m,src,des;int dep[VM], gap[VM];void addedge(int cu, int cv, int cw){ edge[e].frm = cu; edge[e].to = cv; edge[e].cap = cw; edge[e].nxt = head[cu]; head[cu] = e++; edge[e].frm = cv; edge[e].to = cu; edge[e].cap = 0; edge[e].nxt = head[cv]; head[cv] = e++;}int que[VM];void BFS(){ memset(dep, -1, sizeof(dep)); memset(gap, 0, sizeof(gap)); gap[0] = 1; int front = 0, rear = 0; dep[des] = 0; que[rear++] = des; int u, v; while (front != rear) { u = que[front++]; front = front%VM; for (int i=head[u]; i!=-1; i=edge[i].nxt) { v = edge[i].to; if (edge[i].cap != 0 || dep[v] != -1) continue; que[rear++] = v; rear = rear % VM; ++gap[dep[v] = dep[u] + 1]; } }}int cur[VM],stack[VM];int Sap() //sap模板{ int res = 0; BFS(); int top = 0; memcpy(cur, head, sizeof(head)); int u = src, i; while (dep[src] < n) { if (u == des) { int temp = inf, inser = n; for (i=0; i!=top; ++i) if (temp > edge[stack[i]].cap) { temp = edge[stack[i]].cap; inser = i; } for (i=0; i!=top; ++i) { edge[stack[i]].cap -= temp; edge[stack[i]^1].cap += temp; } res += temp; top = inser; u = edge[stack[top]].frm; } if (u != des && gap[dep[u] -1] == 0) break; for (i = cur[u]; i != -1; i = edge[i].nxt) if (edge[i].cap != 0 && dep[u] == dep[edge[i].to] + 1) break; if (i != -1) { cur[u] = i; stack[top++] = i; u = edge[i].to; } else { int min = n; for (i = head[u]; i != -1; i = edge[i].nxt) { if (edge[i].cap == 0) continue; if (min > dep[edge[i].to]) { min = dep[edge[i].to]; cur[u] = i; } } --gap[dep[u]]; ++gap[dep[u] = min + 1]; if (u != src) u = edge[stack[--top]].frm; } } return res;}int main(){ int T, i; scanf("%d", &T); while (T--) { scanf("%d%d", &n, &m); int x, y; int Min = inf, Max = -inf; for (i=1; i<=n; ++i) //找出起点src 终点des { scanf("%d%d", &x, &y); if (x <= Min) { src = i; Min = x; } if (x >= Max) { des = i; Max = x; } } e = 0; memset(head, -1, sizeof(head)); int u, v, c; for (i=0; i!=m; ++i) { scanf("%d%d%d", &u, &v, &c); addedge(u,v,c); addedge(v,u,c); } int ans = Sap(); printf("%d\n", ans); } return 0;}