Drainage Ditches

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

50

题目大意：

      就是因为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，并且是网络水沟，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各自是水沟的起点，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是否已訪问

bool BFS(int s, int t)   //推断是否存在增广路
{
    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) { //仅仅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;  
}