旅游路线问题 线性规划网络流

Posted 2022-12-10 扣得君

旅游路线问题

#include <iostream>在这里插入图片描述

#include <cstring>
#include <map>
#include <queue>
using namespace std;
using std::cout;
const int INF = 1000000;    //正无穷
const int NODESIZE = 100;   //结点最大个数
const int EDGESIZE = 10000; //最大边数
int top;                    //当前边下标
int maxflow;                //最大流最小费用
bool vis[NODESIZE];         //访问标记数组
int c[NODESIZE];            //入队次数
int dist[NODESIZE];         //dist[i]表示源点到点i最短距离:距离即这条路单位cost和
int pre[NODESIZE];          //前驱数组
string str[NODESIZE];
map<string, int> maze;

struct Vertex
              //邻接表头节点
    int first; //与之连接的边的序号
 V[NODESIZE];
struct Edge
                //边表示
    int v, next; //v弧头 next指向下一条邻接边
    int cap, flow, cost;
 E[EDGESIZE];

void init();                                  //初始化
void add(int u, int v, int c, int cost);      //更新混合网络
void add_edge(int u, int v, int c, int cost); //更新混合网络边
void printgraph(int n);                       //输出网络邻接表
void printflow(int n);                        //输出实流边
int MCMF(int s, int t, int n);                //最小花费最大流
bool SPFA(int s, int t, int n);               //求最小费用路
void print(int s, int t);

int main(int argc, char **argv)

    int n, m, i;
    string str1, str2;
    cout << "输入景点个数n和直达路线数m:\\n";
    cin >> n >> m;
    init();
    maze.clear();
    cout << "输入景点名字\\n";
    for (i = 1; i <= n; i++)
    
        cin >> str[i];
        maze[str[i]] = i;
        if (i == 1 || i == n)
        
            add(i, i + n, 2, 0);
        
        else
        
            add(i, i + n, 1, 0);
        
    
    cout << "输入可以直达的两个景点名\\n";
    for (i = 1; i <= m; i++)
    
        cin >> str1 >> str2;
        int a = maze[str1], b = maze[str2];
        if (a < b)
        
            if (a == 1 && b == n)
            
                add(a + n, b, 2, -1);
            
            else
            
                add(a + n, b, 1, -1);
            
        
        else
        
            if (b == 1 && a == n)
            
                add(b + n, a, 2, -1);
            
            else
            
                add(b + n, a, 1, -1);
            
        
    
    cout << "最多经过景点个数:" << 0 - MCMF(1, 2 * n, 2 * n) << endl;
    cout << "依次经过景点:\\n";
    cout << str[1] << endl;
    memset(vis, 0, sizeof(vis));
    print(1, n);
    cout << str[1] << endl;
    return 0;


//初始化
void init()

    memset(V, -1, sizeof(V)); //初始化顶点
    top = 0;                  //当前边下标
    maxflow = 0;


//更新混合网络
void add(int u, int v, int c, int cost)

    add_edge(u, v, c, cost);
    add_edge(v, u, 0, -cost);


//更新混合网络边
void add_edge(int u, int v, int c, int cost)

    //    top    top.v
    //u---------->v
    //构建邻接表：头插法 顺序存储法
    E[top].v = v;
    E[top].cap = c;
    E[top].flow = 0;
    E[top].cost = cost;
    E[top].next = V[u].first; //.next记录链的结点，下一个边的下标
    V[u].first = top++;       //顺序存储拉链

//输出网络邻接表
void printgraph(int n)

    cout << "\\n网络邻接表\\n";
    for (int i = 1; i <= n; i++)
    
        cout << "v" << i << " [" << V[i].first;
        for (int j = V[i].first; ~j; j = E[j].next)
        
            cout << "]--[" << E[j].v << "  " << E[j].cap << "  "
                 << E[j].flow << " " << E[j].cost << " " << E[j].next << "]\\n";
        
        cout << "\\n";
    

//输出实流边
void printflow(int n)

    cout << "实流边:\\n";
    for (int i = 1; i <= n; i++)
    
        for (int j = V[i].first; ~j; j = E[j].next)
        
            if (E[j].flow > 0)
            
                cout << "v" << i << "--"
                     << "v" << E[j].v << " " << E[j].flow << " " << E[j].cost << "\\n";
            
        
    

//最小花费最大流
int MCMF(int s, int t, int n)

    int d; //可增量
    int i, mincost;
    mincost = 0; //maxflow为网络当前最大流量，mincost为网络当前最小费用
    while (SPFA(s, t, n))
                                  //有从s到t的最小费用路
        d = INF;                   //初始化增流量
        cout << "增广路径: " << t; //             i       i^1     i       i+1            i      i+1=i^1   i    i-1=i^1
        for (i = pre[t]; i != -1; i = pre[E[i ^ 1].v])
                                             //i=pre[u.v] u---->v u<----v u---->v u<---v,通俗些就是u-->v就v<---u   v<--u u-->v
            d = min(d, E[i].cap - E[i].flow); //迭代找最小可增量
            cout << "--" << E[i ^ 1].v;
        
        cout << "\\n";
        cout << "增流量: " << d << "\\n";
        //更新最大流
        maxflow += d;
        //增广路上正向边流量+d 反向边流量-d
        for (int i = pre[t]; i != -1; i = pre[E[i ^ 1].v])
        
            E[i].flow += d;
            E[i ^ 1].flow -= d;
        
        mincost += dist[t] * d; //源点到t的单位花费*新增的流量
    
    return mincost;

//求最小费用路
bool SPFA(int s, int t, int n)

    int u, v;
    queue<int> qu;                   //队列
    memset(vis, false, sizeof(vis)); //标记结点是否已经访问过了
    memset(c, 0, sizeof(c));         //入队次数
    memset(pre, -1, sizeof(pre));    //前驱数组初始化为-1
    //距离初始化:源点到各个结点的最短距离
    for (int i = 1; i <= n; i++)
    
        dist[i] = INF;
    
    //源点入队
    vis[s] = true;
    c[s]++;
    dist[s] = 0;
    qu.push(s);

    while (!qu.empty())
    
        //取队头,并消除标记
        u = qu.front();
        qu.pop();
        vis[u] = false;
        //遍历结点u的邻接表：即遍历u的所有出度边u--->x
        for (int i = V[u].first; i != -1; i = E[i].next)
        
            v = E[i].v; //u---->v
            if (E[i].cap > E[i].flow && dist[v] > dist[u] + E[i].cost)
             //松弛操作:这条边还可以增流且借助u-->v比直接到v cost少,如果不可增流则这条边不连通
                //更新源点--->v cost
                dist[v] = dist[u] + E[i].cost;
                //记录v的前驱,pre记录的是边-->v 通过这条边最短到v 则v的前驱为这条边的下标
                pre[v] = i;
                //检测v是否在队列内
                if (!vis[v])
                 //不在
                    //v结点入队列
                    c[v]++;
                    qu.push(v); //入队
                    vis[v] = true;
                    if (c[v] > n)
                     //超过入队上上限，则说明有负环
                        return false;
                    
                
            
        
    
    //最短可增流路径
    cout << "最短可增流路径数组:\\n";
    cout << "dist[]=>";
    for (int i = 1; i <= n; i++)
    
        cout << " " << dist[i];
    
    cout << "\\n";
    if (dist[t] == INF)
     //如果源点到汇点距离为正无穷，则不通:找不出最短可通路径
        return false;
    
    return true;


void print(int s, int t)

	cout<<"s->t:"<<s<<" "<<t<<"  "; 
    int v;
    vis[s] = 1;
    for (int i = V[s].first; ~i; i = E[i].next)
     
        v = E[i].v;
        //(E[i].flow>0&&E[i].cost<=0) 正向路线
        //(E[i].flow<0&&E[i].cost>=0) 反向路线
        if (!vis[v] && ((E[i].flow > 0 && E[i].cost <= 0) || (E[i].flow < 0 && E[i].cost >= 0)))
        
            print(v, t);
            if (v <= t)
            
                cout << str[v] << endl;
            
        
    


/*test
8 10
zhengzhou
luoyang
xian
chengdu
kangding
xianggelila
motuo
lasa
zhengzhou luoyang
zhengzhou xian
luoyang xian
luoyang chengdu
xian chengdu
xian xianggelila
chengdu lasa
kangding motuo
xianggelila lasa
motuo lasa


result:
最短可增流路径数组:
dist[]=> 0 -1 -2 -3 1000000 -3 1000000 -4 0 -1 -2 -3 1000000 -3 1000000 -4
增广路径: 16--8--14--6--11--3--10--2--9--1
增流量: 1
最短可增流路径数组:
dist[]=> 0 0 -1 -1 1000000 -1 1000000 -2 0 0 0 -1 1000000 -1 1000000 -2
增广路径: 16--8--12--4--10--3--9--1
增流量: 1
最短可增流路径数组:
dist[]=> 0 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000
最多经过景点个数:6
依次经过景点:
zhengzhou
luoyang
chengdu
lasa
xianggelila
xian
zhengzhou


*/