标签：des style blog http java color
5 4 1 2 40 1 4 20 2 4 20 2 3 30 3 4 10
50
题目大意：
就是由于下大雨的时候约翰的农场就会被雨水给淹没，无奈下约翰不得不修建水沟，而且是网络水沟，并且聪明的约翰还控制了水的流速，本题就是让你求出最大流速，无疑要运用到求最大流了。题中m为水沟数，n为水沟的顶点，接下来Si,Ei,Ci分别是水沟的起点，终点以及其容量。求源点1到终点n的最大流速。注意重边
</pre></p><pre name="code" class="cpp">
EdmondsKarp算法写的：
邻接矩阵：
</pre><pre name="code" class="cpp">
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#define INF 0x3f3f3f3f
using namespace std;
const int M = 1000 + 50;
int n, m;
int r[M][M];
int pre[M];// 记录结点i的前向结点为pre[i]
bool vist[M];// 记录结点i是否已访问
bool BFS(int s, int t) //判断是否存在增广路
{
queue<int>que;
memset(pre, 0, sizeof(pre));
memset(vist, false, sizeof(vist));
pre[s] = s;
vist[s] = true;
que.push(s);
int p;
while( !que.empty() )
{
p = que.front();
que.pop();
for(int i=1; i<=n; i++)
{
if(r[p][i]>0 && !vist[i])
{
pre[i]=p;
vist[i]=true;
if( i==t )
return true;
que.push(i);
}
}
}
return false;
}
int EK(int s, int t)
{
int maxflow = 0;
while( BFS(s, t) )
{
int d = INF;
// 若有增广路径，则找出最小的delta
for(int i=t; i!=s; i=pre[i])
d = min(d, r[ pre[i] ][i]);
// 这里是反向边
for(int i=t; i!=s; i=pre[i])
{
r[ pre[i] ][i] -= d;//方向边
r[i][ pre[i] ] += d;//方向边
}
maxflow += d;
}
return maxflow;
}
int main()
{
while(cin>>m>>n)
{
memset(r, 0, sizeof(r));
for(int i=0; i<m; i++)
{
int from, to, rap;
scanf("%d%d%d", &from, &to, &rap);
r[from][to] += rap;
}
cout<<EK(1, n)<<endl;
}
return 0;
}
邻接表（紫书上的模板）：
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f;
const int MAXN = 1000 + 50;
struct Edge
{
int from, to, cap, flow;
Edge (int u, int v, int c, int f):from(u), to(v), cap(c), flow(f) {}
};
struct EdmondsKarp
{
int n, m;
vector<Edge> edges;
vector<int> G[MAXN];
int a[MAXN];
int p[MAXN];
void init(int n)
{
for(int i=0; i<n; i++)
G[i].clear();
edges.clear();
}
void AddEdge(int from, int to, int cap)
{
edges.push_back( Edge(from, to, cap, 0) );
edges.push_back( Edge(to, from, 0, 0) );
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
int Maxflow(int s, int t)
{
int flow = 0;
for( ; ; )
{
memset(a, 0, sizeof(a));
queue<int> Q;
Q.push(s);
a[s]=INF;
while( !Q.empty() )
{
int x = Q.front();
Q.pop();
for(int i=0; i<G[x].size(); i++)
{
Edge& e = edges[ G[x][i] ];
if( !a[e.to] && e.cap > e.flow )
{
p[e.to] = G[x][i];
a[e.to] = min(a[x], e.cap-e.flow);
Q.push(e.to);
}
}
if( a[t] ) break;
}
if( ! a[t] ) break;
for(int u=t; u!=s; u=edges[ p[u] ].from )
{
edges[ p[u] ].flow += a[t];
edges[ p[u]^1 ].flow -= a[t];
}
flow += a[t];
}
return flow;
}
};
int main()
{
EdmondsKarp T;
int n, m;
while(scanf("%d%d", &m, &n) !=EOF)
{
T.init(n+1);
for(int i=0; i<m; i++)
{
int a1, a2, a3;
scanf("%d%d%d", &a1, &a2, &a3);
T.AddEdge(a1, a2, a3);
}
printf("%d\n", T.Maxflow(1, n));
}
return 0;
}
Dinic算法：
#include <cstdio>
#include <cstring>
#include <queue>
#define MAXN 205
#define INF 1000000000
using namespace std;
struct Edge {
int from, to, cap, flow;
};
struct Dinic {
int n, m, s, t;
vector<Edge> edges; //边表.edges[e]和edges[e^1]互为反向弧
vector<int> G[MAXN]; //邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
bool vis[MAXN]; //BFS使用
int d[MAXN]; //从起点到i的距离
int cur[MAXN]; //当前弧指针
void ClearAll(int n) {
for (int i = 0; i < n; i++) G[i].clear();
edges.clear();
}
void AddEdge(int from, int to, int cap) {
edges.push_back((Edge) {from, to, cap, 0});
edges.push_back((Edge) {to, from, 0, 0});
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool BFS() {//使用BFS计算出每一个点在残量网络中到t的最短距离d.
memset(vis, 0, sizeof(vis));
queue<int> Q;
Q.push(s);
vis[s] = 1;
d[s] = 0;
while (!Q.empty()) {
int x = Q.front(); Q.pop();
for (int i = 0; i < G[x].size(); i++) {
Edge& e = edges[G[x][i]];
if (!vis[e.to] && e.cap > e.flow) { //只考虑残量网络中的弧
vis[e.to] = 1;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x, int a) {//使用DFS从S出发，沿着d值严格递减的顺序进行多路增广。
if (x == t || a == 0) return a;
int flow = 0, f;
for (int& i = cur[x]; i < G[x].size(); i++) {
Edge& e = edges[G[x][i]];
if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[G[x][i] ^ 1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
}
int Maxflow(int s, int t) {
this->s = s; this->t = t;
int flow = 0;
while (BFS()) {
memset(cur, 0, sizeof(cur));
flow += DFS(s, INF);
}
return flow;
}
};
Dinic g;
int main()
{
int n, m, i, a, b, c;
while (~scanf("%d%d", &m, &n)) {
g.ClearAll(n + 1);
for (i = 0; i < m; i++) {
scanf("%d%d%d", &a, &b, &c);
g.AddEdge(a, b, c);
}
int flow = g.Maxflow(1, n);
printf("%d\n", flow);
}
return 0;
}
HDU 1532||POJ1273：Drainage Ditches（最大流）,布布扣,bubuko.com
HDU 1532||POJ1273：Drainage Ditches（最大流）
标签：des style blog http java color
原文地址：http://blog.csdn.net/u013487051/article/details/37699643
