USACO 3.2 ratios 高斯消元

  1 /*
  2 PROB:ratios
  3 LANG:C++
  4 */
  5 
  6 #include <stdio.h>
  7 #include <math.h>
  8 #include <algorithm>
  9 #include <iostream>
 10 using namespace std;
 11 #define maxn 5
 12 double A[5][5],a[5][5];
 13 
 14 typedef double Matrix[maxn][maxn];
 15 
 16 void solve(Matrix A,int n)
 17 {
 18     int i,j,k,r;
 19 
 20     for (i=0;i<n;i++)
 21     {
 22         r=i;
 23         for (j=i+1;j<n;j++)
 24             if (fabs(A[j][i])>fabs(A[r][i]))
 25                 r=j;
 26         if (r!=i)
 27             for (j=0;j<=n;j++)
 28                 swap(A[r][j],A[i][j]);
 29 
 30         for (k=i+1;k<n;k++)
 31         {
 32             double f=A[k][i]/A[i][i];
 33             for (j=i;j<=n;j++)
 34                 A[k][j]-=f*A[i][j];
 35         }
 36     }
 37     for (i=n-1;i>=0;i--)
 38     {
 39         for (j=i+1;j<n;j++)
 40             A[i][n]-=A[j][n]*A[i][j];
 41         A[i][n]/=A[i][i];
 42     }
 43 }
 44 
 45 bool satisify(double x,double y,double z)
 46 {
 47     int t1=x+0.5,t2=y+0.5,t3=z+0.5;     //+0.5四舍五入，处理精度问题
 48                                         //比如原来是4.9999999，直接转成int就成了4，完蛋啦T^T
 49     double xx=x-t1,yy=y-t2,zz=z-t3;
 50     double comp=pow(10,-5);             //double的比较法= =，差值小于10^-10就算相等了
 51     if ((fabs(xx)<=comp)&&(fabs(yy)<=comp)&&(fabs(zz)<=comp))
 52         return true;
 53     else
 54         return false;
 55 }
 56 
 57 int main()
 58 {
 59     freopen("ratios.in","r",stdin);
 60     freopen("ratios.out","w",stdout);
 61 
 62     int x,y,z,t1,t2,t3;
 63     double dx,dy,dz;
 64     bool sol=false;
 65 
 66     scanf("%d %d %d",&x,&y,&z);
 67     //A[0][3]=x;  A[1][3]=y;  A[2][3]=z;
 68     t1=x;   t2=y;   t3=z;
 69     scanf("%d %d %d",&x,&y,&z);
 70     A[0][0]=x;  A[1][0]=y;  A[2][0]=z;
 71     scanf("%d %d %d",&x,&y,&z);
 72     A[0][1]=x;  A[1][1]=y;  A[2][1]=z;
 73     scanf("%d %d %d",&x,&y,&z);
 74     A[0][2]=x;  A[1][2]=y;  A[2][2]=z;
 75 /*
 76     for (int i=0;i<=2;i++)
 77     {
 78         for (int j=0;j<=3;j++)
 79             printf("%.8f ",A[i][j]);
 80         printf("\n");
 81     }
 82 */
 83     for (int i=1;i<=100;i++)
 84     {
 85         for (int i=0;i<=2;i++)
 86             for (int j=0;j<=3;j++)
 87                 a[i][j]=A[i][j];
 88         a[0][3]=t1*i;
 89         a[1][3]=t2*i;
 90         a[2][3]=t3*i;
 91         //printf("%.8f %.8f %.8f\n",a[0][3],a[1][3],a[2][3]);
 92         solve(a,3);
 93         dx=a[0][3];     dy=a[1][3];     dz=a[2][3];
 94         //printf("%.8f %.8f %.8f\n",dx,dy,dz);
 95         if (satisify(dx,dy,dz))
 96         {
 97             sol=true;
 98             int xx=dx,yy=dy,zz=dz;
 99             //printf("%d %d %d\n",xx,yy,zz);
100             if (xx<0||yy<0||zz<0)
101                 printf("NONE\n");
102             else
103             {
104                 xx=dx+0.5;  yy=dy+0.5;  zz=dz+0.5;
105                 printf("%d %d %d %d\n",xx,yy,zz,i);
106             }
107             break;
108         }
109         //printf("%.8f %.8f %.8f\n",a[0][3],a[1][3],a[2][3]);
110     }
111     if (!sol) printf("NONE\n");
112 
113 }
114 
115 /*
116 
117 
118 int main()              //垃圾代码，一开始YY的错了
119 {
120     freopen("ratios.in","r",stdin);
121     freopen("ratios.out","w",stdout);
122 
123     double a1,a2,a3,a0,b1,b2,b3,b0,c1,c2,c3,c0,A0,A1,A2,A3,B0,B1,B2,B3,C0,C1,C2,C3;
124     int xx,yy,zz;
125     bool sol=false;
126     cin>>A0>>B0>>C0;
127     cin>>A1>>B1>>C1;    //line1
128     cin>>A2>>B2>>C2;    //line2
129     cin>>A3>>B3>>C3;    //line3
130 
131 for (int i=1;i<=100;i++)
132 {
133     a0=A0*i;  a1=A1;  a2=A2;  a3=A3;
134     b0=B0*i;  b1=B1;  b2=B2;  b3=B3;
135     c0=C0*i;  c1=C1;  c2=C2;  c3=C3;
136 
137     double t0=b1*c1,t1=c1*a1,t2=a1*b1;
138     a1=a1*t0;   a2=a2*t0;   a3=a3*t0;   a0=a0*t0;
139     b1=b1*t1;   b2=b2*t1;   b3=b3*t1;   b0=b0*t1;
140     c1=c1*t2;   c2=c2*t2;   c3=c3*t2;   c0=c0*t2;
141     c1=c1-a1;   c2=c2-a2;   c3=c3-a3;   c0=c0-a0;
142     b1=b1-a1;   b2=b2-a2;   b3=b3-a3;   b0=b0-a0;
143     t0=c2;      t1=b2;
144     b1=b1*t0;   b2=b2*t0;   b3=b3*t0;   b0=b0*t0;
145     c1=c1*t1;   c2=c2*t1;   c3=c3*t1;   c0=c0*t1;
146     c1=c1-b1;   c2=c2-b2;   c3=c3-b3;   c0=c0-b0;
147     double z=c0/c3;
148     double y=(b0-b3*z)/b2;
149     double x=(a0-a3*z-a2*y)/a1;
150     printf("%%f %f %f\n",x,y,z);
151     if (satisify(x,y,z))
152     {
153         sol=true;
154         xx=x,yy=y,zz=z;
155         if (xx<0||yy<0||zz<0)
156             printf("NONE\n");
157         else
158             printf("%d %d %d %d\n",xx,yy,zz,i);
159         break;
160     }
161 }
162     if (!sol) printf("NONE\n");
163     return 0;
164 }
165 */