图论网络流模板

最大流
O ( n 2 v ) O(n^2*\sqrt{v})

//isap

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
typedef long long LL;
const int MAXN = 5010;//点数的最大值
const int MAXM = 1e7+10;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw = 0)
{
    edge[tol] = Edge{v,head[u],w,0};
    head[u] = tol++;
    edge[tol] =Edge{u,head[v],rw,0};
    head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0] = 1;
    int front = 0, rear = 0;
    dep[end] = 0;
    Q[rear++] = end;
    while(front != rear)
    {
        int u = Q[front++];
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(dep[v] != -1)continue;
            Q[rear++] = v;
            dep[v] = dep[u] + 1;
            gap[dep[v]]++;
        }
    }
}
int S[MAXN];
int sap(int start,int end,int N)
{
    BFS(start,end);
    memcpy(cur,head,sizeof(head));
    int top = 0;
    int u = start;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            int inser;
            for(int i = 0; i < top; i++)
                if(Min > edge[S[i]].cap - edge[S[i]].flow)
                {
                    Min = edge[S[i]].cap - edge[S[i]].flow;
                    inser = i;
                }
            for(int i = 0; i < top; i++)
            {
                edge[S[i]].flow += Min;
                edge[S[i]^1].flow -= Min;
            }
            ans += Min;
            top = inser;
            u = edge[S[top]^1].to;
            continue;
        }
        bool flag = false;
        int v;
        for(int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
            {
                flag = true;
                cur[u] = i;
                break;
            }
        }
        if(flag)
        {
            S[top++] = cur[u];
            u = v;
            continue;
        }
        int Min = N;
        for(int i = head[u]; i != -1; i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]])return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if(u != start)u = edge[S[--top]^1].to;
    }
    return ans;
}

const int MAXN = 100010;//点数的最大值
const int MAXM = 400010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init()
{
    tol = 0;
    memset(head,-1,sizeof(head));
}
//加边,单向图三个参数,双向图四个参数
void addedge(int u,int v,int w,int rw = 0)
{
    edge[tol] = Edge{v,head[u],w,0};
    head[u] = tol++;
    edge[tol] =Edge{u,head[v],rw,0};
    head[v] = tol++;
}
//输入参数:起点、终点、点的总数
//点的编号没有影响,只要输入点的总数
int sap(int start,int end,int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            for(int i = pre[u]; i != -1; i = pre[edge[i^1].to])
                if(Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for(int i = pre[u]; i != -1; i = pre[edge[i^1].to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        bool flag = false;
        int v;
        for(int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
            {
                flag = true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if(flag)
        {
            u = v;
            continue;
        }
        int Min = N;
        for(int i = head[u]; i != -1; i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]])return ans;
        dep[u] = Min+1;
        gap[dep[u]]++;
        if(u != start) u = edge[pre[u]^1].to;
    }
    return ans;
}

邻接矩阵,可以跑多次,不用拆图,写起来简单

const int MAXN=110;
int maze[MAXN][MAXN];
int gap[MAXN],dis[MAXN],pre[MAXN],cur[MAXN];
int flow[MAXN][MAXN];//存最大流的容量
int sap(int start,int end,int nodenum)
{
    memset(cur,0,sizeof(cur));
    memset(dis,0,sizeof(dis));
    memset(gap,0,sizeof(gap));
    memset(flow,0,sizeof(flow));
    int u=pre[start]=start,maxflow=0,aug=-1;
    gap[0]=nodenum;
    while(dis[start]<nodenum)
    {
        loop:
        for(int v=cur[u]; v<nodenum; v++)
            if(maze[u][v]-flow[u][v] && dis[u]==dis[v]+1)
            {
                if(aug==-1 ||
                        aug>maze[u][v]-flow[u][v])aug=maze[u][v]-flow[u][v];
                pre[v]=u;
                u=cur[u]=v;
                if(v==end)
                {
                    maxflow+=aug;
                    for(u=pre[u]; v!=start; v=u,u=pre[u])
                    {
                        flow[u][v]+=aug;
                        flow[v][u]-=aug;
                    }
                    aug=-1;
                }
                goto loop;
            }
        int mindis=nodenum-1;
        for(int v=0; v<nodenum; v++)
            if(maze[u][v]-flow[u][v]&&mindis>dis[v])
            {
                cur[u]=v;
                mindis=dis[v];
            }
        if((--gap[dis[u]])==0)break;
        gap[dis[u]=mindis+1]++;
        u=pre[u];
    }
    return maxflow;
}

最小费用最大流

玄学复杂度


#include<iostream>
#include<stdio.h>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 1000;
typedef long long LL;
typedef pair<int,int> PII;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge{
    int to, next, cap, flow, cost;
} edge[MAXM];
int head[MAXN], tol;
int pre[MAXN], dis[MAXN];
bool vis[MAXN];
int N;                  //0~N-1的点数
void init(int n)
{
    N = n;
    tol = 0;
    memset(head, -1, sizeof(head));
}

void addedge(int u, int v, int cap, int cost)
{
    edge[tol] = Edge{v,head[u],cap,0,cost};
    head[u] = tol++;
    edge[tol] = Edge{u,head[v],0,0,-cost};
    head[v] = tol++;
}

bool spfa(int s, int t){
    queue<int> q;
    for (int i = 0; i < N; i++){
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while (!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = false;
        for (int i = head[u]; i != -1; i = edge[i].next){
            int v = edge[i].to;

            if (edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost){
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if (!vis[v]){
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if (pre[t] == -1) return false;
    else  return true;
}

int minCostMaxflow(int s, int t, int &cost){
    int flow = 0;
    cost = 0;
    while (spfa(s, t)){
        int Min = INF;
        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]){
            if (Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to]){
            edge[i].flow += Min;
            edge[i ^ 1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
    return flow;
}


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转载自blog.csdn.net/wyxxzsy/article/details/82980894
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