C++ Dijkstra::init方法代码示例

int main() {
  int r1, c1, r2, c2, kase = 0;
  while(scanf("%d%d%d%d%d%d", &R, &C, &r1, &c1, &r2, &c2) == 6 && R) {
    r1--; c1--; r2--; c2--;
    for(int r = 0; r < R; r++) {
      for(int c = 0; c < C-1; c++)
        grid[r][c][RIGHT] = grid[r][c+1][LEFT] = readint();
      if(r != R-1) for(int c = 0; c < C; c++)
        grid[r][c][DOWN] = grid[r+1][c][UP] = readint();
    }
    solver.init(R*C*5+1);

    n = 0;
    memset(id, 0, sizeof(id));

    // 源点出发的边
    for(int dir = 0; dir < 4; dir++) if(cango(r1, c1, dir)) {
       solver.AddEdge(0, ID(r1+dr[dir], c1+dc[dir], dir), grid[r1][c1][dir]*2); // 开始走下去
       solver.AddEdge(0, ID(r1+dr[dir], c1+dc[dir], 4), grid[r1][c1][dir]*2); // 走一步停下来
    }

    // 计算每个状态(r,c,dir)的后继状态
    for(int r = 0; r < R; r++)
      for(int c = 0; c < C; c++) {
        for(int dir = 0; dir < 4; dir++) if(cango(r, c, inv[dir])) {
          solver.AddEdge(ID(r, c, dir), ID(r, c, 4), grid[r][c][inv[dir]]); // 停下来！
          if(cango(r, c, dir))
            solver.AddEdge(ID(r, c, dir), ID(r+dr[dir], c+dc[dir], dir), grid[r][c][dir]); // 继续走
        }
        for(int dir = 0; dir < 4; dir++) if(cango(r, c, dir)) {
          solver.AddEdge(ID(r, c, 4), ID(r+dr[dir], c+dc[dir], dir), grid[r][c][dir]*2); // 重新开始走
          solver.AddEdge(ID(r, c, 4), ID(r+dr[dir], c+dc[dir], 4), grid[r][c][dir]*2); // 走一步停下来
        }
      }

    // 跑dijkstra
    solver.dijkstra(0);

    // 找最优解
    int ans = solver.d[ID(r2, c2, 4)];

    printf("Case %d: ", ++kase);
    if(ans == INF) printf("Impossible\n"); else printf("%d\n", ans);
  }
  return 0;
}

int main() {
  int kase = 0, N, S, E, M, K, X, Y, Z;
  while(scanf("%d%d%d%d", &N, &S, &E, &M) == 4) {
    solver.init(N);

    S--; E--; // 编号从0~N-1
    for(int i = 0; i < M; i++) {
      scanf("%d%d%d", &X, &Y, &Z); X--; Y--;
      solver.AddEdge(X, Y, Z);
      solver.AddEdge(Y, X, Z);
    }
    solver.GetShortestPaths(S, d1, paths1); // S到所有点的距离和路径
    solver.GetShortestPaths(E, d2, paths2); // T到所有点的距离和路径

    int ans = d1[E];              // 初始解解为直达距离
    vector<int> path = paths1[E]; // 初始解的station序列
    int midpoint = -1;            // 不坐商业线

    scanf("%d", &K);
    for(int i = 0; i < K; i++) {
      scanf("%d%d%d", &X, &Y, &Z); X--; Y--;
      for(int j = 0; j < 2; j++) { // j=0代表商业线坐X->Y，j=1代表Y->X
        if(d1[X] + d2[Y] + Z < ans) {
          ans = d1[X] + d2[Y] + Z;
          path = paths1[X];
          for(int j = paths2[Y].size()-1; j >= 0; j--) // 从Y到T的距离要反过来
            path.push_back(paths2[Y][j]);
          midpoint = X;
        }
        swap(X, Y);
      }
    }

    if(kase != 0) printf("\n");
    kase++;

    for(int i = 0; i < path.size()-1; i++) printf("%d ", path[i]+1);
    printf("%d\n", E+1);
    if(midpoint == -1) printf("Ticket Not Used\n"); else printf("%d\n", midpoint+1);
    printf("%d\n", ans);
  }
  return 0;
}

int main() 
{
  int n, m;
  while(scanf("%d%d", &n, &m) == 2) 
  {
    solver.init(n);
    for(int i = 0; i < m; ++i) 
	{
      int a, b, c;
      scanf("%d%d%d", &a, &b, &c); --a; --b;
      solver.AddEdge(a, b, c);
      solver.AddEdge(b, a, c);
    }

    solver.dijkstra(1); // 家(1)到所有点的距离。因为道路都是双向的，所以把家看作起点也行
    memset(d, -1, sizeof(d));
    printf("%d\n", dp(0)); // 办公室(0)到家的符合条件的路径条数
  }
  return 0;
}

int main()
{
    int t;
    scanf("%d",&t);
    for(int i = 1;i <= t;i ++)
    {
        memset(edges,0,sizeof(edges));
        memset(next,0,sizeof(next));
        memset(head,0,sizeof(head));
        maps.clear();
        int tott = 0;
        tot = 0;
        scanf("%d",&n);
        dijkstra.init(n);
        for(int i = 1;i <= n;i ++)
        {
            string s1;
            int ct;
            cin>>s1;
            maps[s1] = ++tott;
            scanf("%d",&ct);
            for(int i = 1;i <= ct;i ++)
            {
                int a,b;
                scanf("%d%d",&a,&b);
                build(maps[s1],a,b);
            }
        }
        int q;
        scanf("%d",&q);
        string s1,s2;
        for(int i = 1;i <= q;i ++)
        {
            cin>>s1>>s2;
            dijkstra.dijkstra(maps[s1]);
            printf("%d\n",dijkstra.d[maps[s2]]);
        }
    }
    return 0;
}

int main(int argc, char const *argv[])
{
	READ;
	WRITE;
	read();
	Dijkstra dijkstra;
	dijkstra.init();
	for(int i = 0 ; i < 300 ; i ++)
	{
		for(int j = 0 ; j < 300 ; j++)
		{
			if(g[i][j]!=-1)
			{
				// printf("* %c %c %d\n",i,j,g[i][j]);
				dijkstra.AddEdge(i, j, g[i][j]);
			}
		}
	}

	dijkstra.dijkstra('Z');
	return 0;
}