[JSOI2007] Alloy

Title Description

A company processes an alloy consisting of iron, aluminum, and tin. Their work is very simple. First import some Fe-al-SN alloy raw materials, different kinds of raw materials in the proportion of Fe-al-Sn. Then, each raw material is taken out of a certain amount, after melting, mixing, to obtain a new alloy. The new alloy's Fe-al-SN specific gravity is required by the user.

Now, the user gives the n kinds of alloys they need, and the proportion of Fe-al-sn in each alloy. The company wants to be able to order a minimum variety of raw materials and use the raw materials to produce all the kinds of alloys the user needs.

Input/output format

Input format:

The first line is two integers m and n (M, n≤500), which represent the number of raw materials and the number of alloy species required by the user, respectively.

2nd to M + 1 lines, each line three real numbers a, B, C (A, B, c≥0 and A + b + c = 1), respectively, represents the proportion of iron and aluminum in a raw material.

The M + 2 to M +n + 1 lines, three real numbers a, B, C (A, B, c≥0 and A + b + c = 1) per line, respectively, representing the Fe-al-Sn in a user-required alloy

The proportion of the total.

Output format:

An integer that represents the minimum number of raw material species required. If there is no solution, the output –1.

Input and Output Sample input example # #:

10 100.1 0.2 0.70.2 0.3 0.50.3 0.4 0.30.4 0.5 0.10.5 0.1 0.40.6 0.2 0.20.7 0.3 00.8 0.1 0.10.9 0.1 01 0 00.1 0.2 0.70.2 0. 3 0.50.3 0.4 0.30.4 0.5 0.10.5 0.1 0.40.6 0.2 0.20.7 0.3 00.8 0.1 0.10.9 0.1 01 0 0

Did not resist and went to Kang a computational geometry ...
  First of all it is not difficult to find that C is an unused amount,, because a+b+c=1, so c=1-a-b, when a, B are satisfied when C is certainly satisfied.
So just consider a A-B.
  And because a vector that can be linearly combined by a number of points in a plane must be within the convex hull of these points in the plane, the title becomes
The minimum number of points allows them to form a convex hull that contains all the demand points.

  This is also equivalent to finding a loop starting from one point, so that the demand points are on the left side of the loop and the number of sides on the road is minimized.
So build a picture, run a floyed.

#include <bits/stdc++.h>#definell Long Long#defineMAXN 1005using namespacestd;Const Doubleeps=0.000000001;Doublea,b,c;intN,M,DIS[MAXN][MAXN];structnode{Doublex, y; Nodeoperator-(Constnode& u)Const{        return(node) {x-u.x,y-U.Y}; }}Base[Maxn],user[maxn];inlineintZtDoublex) {    if(Fabs (x) <=eps)return 0; returnX>0?1:-1;} InlineDoublePointmul (node X,node y) {returnx.x*y.x+x.y*y.y;} InlineDoubleXmul (node X,node y) {returnx.x*y.y-y.x*x.y;}intMain () {scanf ("%d%d",&n,&m);  for(intI=1; i<=n+m;i++) {scanf ("%LF%LF%LF",&a,&b,&c); if(i<=n)Base[i]=(node) {A, b}; Elseuser[i-n]=(node) {A, b}; } memset (Dis,0x3f,sizeof(DIS));//for (int i=1;i<=n;i++) dis[i][i]=0;         for(intI=1; i<=n;i++)         for(intj=1; j<=n;j++){            BOOLflag=1; Node Vec=Base[j]-Base[i];  for(intk=1; k<=m;k++){                DoubleX=zt (Xmul (vec,user[k]-Base[i])); if(x<0|| (! X&&zt (Pointmul (user[k]-Base[i],user[k]-Base[j])) >0) ) {flag=0;  Break; }            }                        if(flag) dis[i][j]=1; }        intans=1<< -;  for(intk=1; k<=n;k++)         for(intj=1; j<=n;j++)             for(intI=1; i<=n;i++) Dis[j][i]=min (dis[j][i],dis[j][k]+Dis[k][i]);  for(intI=1; i<=n;i++) ans=min (ans,dis[i][i]); if(Ans>n) puts ("-1"); Elseprintf"%d\n", ans); return 0;}

[JSOI2007] Alloy