Drainage Ditches---hdu1532（最大流, 模板）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1532

最大流模板题；

EK:（复杂度为n*m*m）;

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
#define INF 0xfffffff
#define N 220
int maps[N][N], pre[N], ans;
bool bfs(int s, int e)
{
    memset(pre, 0, sizeof(pre));
    queue<int>Q;
    Q.push(s);
    while(Q.size())
    {
        int i = Q.front();
        Q.pop();
        if(i == e)
            return true;
        for(int j=1; j<=e; j++)
        {
            if(pre[j]==0 && maps[i][j] > 0)
            {
                pre[j] = i;
                Q.push(j);
            }
        }
    }
    return false;
}
void EK(int s, int e)
{
    while(bfs(s, e))
    {
        int Min = INF;
        for(int i=e; i!=s; i=pre[i])
            Min=min(maps[pre[i]][i], Min);
         for(int i=e; i!=s; i=pre[i])
         {
             maps[pre[i]][i]-=Min;
             maps[i][pre[i]]+=Min;
         }
         ans+=Min;
    }
}
int main()
{
    int n, m, x, y,c;
    while(scanf("%d%d", &m, &n)!=EOF)
    {
        memset(maps, 0, sizeof(maps));
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d", &x, &y, &c);
            maps[x][y] += c;
        }
        ans = 0;
        EK(1, n);
        printf("%d
", ans);
    }
    return 0;
}

Dinic:(复杂度为n*n*m)

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;

#define N 220
#define INF 0xfffffff

int n, ans, Head[N], cnt, Layer[N];
struct Edge
{
    int v, flow, next;
} e[2*N];

void Add(int u, int v, int flow)
{
    e[cnt].v = v;
    e[cnt].flow = flow;
    e[cnt].next = Head[u];
    Head[u] = cnt++;
}
bool bfs(int S, int E)
{
    memset(Layer, 0, sizeof(Layer));
    Layer[S] = 1;
    queue<int>Q;
    Q.push(S);
    int p, q;
    while(!Q.empty())
    {
        p = Q.front();
        Q.pop();
        if(p == E)return true;
        for(int i=Head[p]; i!=-1; i=e[i].next)
        {
            q = e[i].v;
            if(!Layer[q] && e[i].flow)
            {
                Layer[q] = Layer[p]+1;
                Q.push(q);
            }
        }
    }
    return false;
}
int dfs(int u, int MaxFlow, int E)
{
    if(u == E)return MaxFlow;
    int  uflow=0;
    for(int i=Head[u]; i!=-1; i=e[i].next)
    {
        int v = e[i].v;
        if(Layer[v]==Layer[u]+1 && e[i].flow)
        {
            int flow = min(e[i].flow, MaxFlow - uflow);
            flow = dfs(v, flow, E);

            e[i].flow -= flow;
            e[i^1].flow += flow;
            uflow += flow;
            if(uflow==MaxFlow)break;
        }
    }
    if(uflow==0)
        Layer[u]=0;
    return uflow;
}
void Dinic()
{
    while(bfs(1, n))
    {
        ans+=dfs(1, INF, n);
    }
}
int main()
{
    int a, b, flow, m;
    while(scanf("%d%d", &m, &n)!=EOF)
    {
        memset(Head, -1, sizeof(Head));
        cnt = 0;
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d", &a, &b, &flow);
            Add(a, b, flow);
            Add(b, a, 0);
        }
        ans = 0;
        Dinic();
        printf("%d
", ans);
    }
    return 0;
}