Might just be a matter of restricting the plot x-y values. I base the selection predicate on values I see in the plot you show.

newpts = Map[Reverse, allres];
newpts2 = Select[newpts, (#[[2]] - 1) + (#[[1]] - 34)/11 <= 0 &];

Now plot it. I have the y coordinate start at 0.05 to remove some seeming badness at the bottom.

ListContourPlot[newpts2, GridLines -> Automatic, 
 ContourLabels -> True, ColorFunction -> "Rainbow", 
 ImageSize -> Large, PlotRange -> {{0, 45}, {.05, 1}}, 
 PlotLegends -> Automatic]

Thanks, Daniel Lichtblau!

I want a plot which should be the same as shown in the figure, which is generated from a different software.

Is there any option in Mathematica to generate the contour plot cleanly.

I don't know that they qualify as jittery, exactly. Try ListPlot3D[Map[Reverse, allres]] to see what I mean. They really show a precipitous change I would describe as a fold, with some jaggedness on one side.

Some trial and error indicates that smoothing the middle coordinates can make a difference to the outcome. I show this below. I use a (fairly long) triangle-type filter to convolve those values.

offset = 17;
ker0 = Join[Range[offset], Range[offset - 1, 1, -1]];
ker = ker0/Total[ker0];
newy = ListConvolve[ker, allres[[All, 2]]];
{newx, newz} = Transpose@allres[[offset ;; -offset, 3 ;; 1 ;; -2]];
newpts = Transpose[{newx, newy, newz}];

ListPlot3D[newpts]

Now the contour plot.

ListContourPlot[newpts, GridLines -> Automatic, ContourLabels -> True,
  ColorFunction -> "Rainbow", ImageSize -> Large]

Still a bit of weirdness at the lower right but much of the jaggedness is gone. One can experiment with the kernel size to get a trade-off between smoothness and fidelity to the original.