两点之间的所有最短路径

---

#include "iostream"
#include "string"
#include "fstream"
#include "vector"
#include "queue"
#include "sstream"
#include "set"
#include "string.h"
#include "math.h"
#include 

using namespace std;

#define INF 200000000
#define N 1000

struct Edge
{
    int to;          // 边终止节点
    int cost;        // 花费

    Edge(int to1, int cost1)
    {
        to = to1;
        cost = cost1;
    }
};

int nV;                      // 顶点数
int nE;                      // 边数
vector G[N];           // 图的邻接表形式
int G1[N][N];                // 图的邻接矩阵形式
int dist[N];                 // 从源点出发的最短距离
typedef pair<int, int> P;    // first是最短距离，second是顶点编号
bool vis[N];
vector G4[N];

void build();                                                  // 建图
void dijkstra(int s, vector G[N], int *preV);            // 求最短路径
void solve();                                                  // 主体


void addEdge(int from, int to, int cost, vector G[N])
{
    Edge e(to, cost);
    G[from].push_back(e);

    Edge e1(from, cost);  
    G[to].push_back(e1);
}

void build()
{
    int i;
    ifstream fin;
    fin.open("data.txt");
    cout << "顶点数：";
    fin >> nV; 
    cout << nV << endl;
    cout << "边数：";
    fin >> nE; 
    cout << nE << endl;

    // 输入图
    for(i=0; ifor(int j=i; jcout << endl << "原图{边的起点，终点，花费}："  << endl;
    int from, to, cost;
    for(i=0; i> from >> to >> cost;
        cout << from << " " << to << " " << cost << endl;
        addEdge(from, to, cost, G);
        G1[from][to] = G1[to][from] = cost;
    }
    fin.close();
}

void dijkstra(int s, vector G[N])
{
    fill(dist, dist + nV+1, INF);
    priority_queuevector
, greater
 > q;
    dist[s] = 0;
    q.push(P(0, s));
    while(!q.empty())
    {
        P p = q.top();   //从尚未使用的顶点中找到一个距离最小的顶点
        q.pop();
        int v = p.second;
        if(dist[v] < p.first)
            continue;
        for(int i=0; iint dis = dist[v] + e.cost;
            if(dist[e.to] > dis)
            {
                dist[e.to] = dist[v] + e.cost;
                q.push(P(dist[e.to], e.to));
                G4[v].push_back(e);
            }
            else if(dist[e.to] == dis)
            {
                G4[v].push_back(e);
            }
        }
    }
}

struct Ans
{
    vector<int> path;
    int cost;
    int start;

    void getCost()
    {
        cost = G1[start][path[0]];
        for(int i=0; i1; i++)
        {
            cost += G1[path[i]][path[i+1]];
        }
    }
};

void dfs(int s, int t, Ans &A, vector< Ans > &paths, int start)
{
    if (s == t)
    {
        A.start = start;
        A.getCost();
        paths.push_back(A);
    }

    for (int i = 0; i < G4[s].size(); i++)
    {
            int u = G4[s][i].to;
            if (!vis[u])
            {
                vis[u] = true;
                A.path.push_back(u);
                dfs(u, t, A, paths, start);
                A.path.pop_back();
                vis[u] = false;
            }
    }
}

void solve()
{
    build();
    dijkstra(1, G);

    int i, j;
    cout << endl << "最短路径图{边的起点，终点，话费}："  << endl;
    for(i=0; ifor(int j=0; jcout << i << " " << G4[i][j].to << " " << G4[i][j].cost << endl;
        }
    }

    vector paths;
    Ans ans;
    memset(vis, false, sizeof(vis));
    dfs(1, 9, ans, paths, 1);

    cout << endl << "1到9的所有最短路径：" << endl;
    for(i=0; icout << "1 ";
        for(j=0; jcout << paths[i].path[j] << " ";
        }
        cout << "---cost：" << paths[i].cost << endl;
    }
}

int main()
{
    solve();
    getchar();
    return 0;
}

参考文献
https://wenku.baidu.com/view/0254244b001ca300a6c30c22590102020740f2d8.html
王志坚，韩伟一，李一军. 具有多条最短路径的最短路问题[J]. 哈尔滨工业大学学报，2010，42（9）：1429-1431.

附
data.txt

9
13
1 2 18
1 3 4
1 4 5
2 4 13
2 9 2
3 4 1
3 6 6
4 5 10
4 6 5
5 9 5
6 7 2
7 8 2
8 9 6