HDU 5095Linearization of the kernel functions in SVM（水题）

最新推荐文章于 2018-05-16 19:36:48 发布

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本文介绍了一种将SVM中的复杂二次核函数通过引入新变量进行线性化的方法，并提供了一个转换程序来实现这一过程。适用于支持向量机在聚类分析等场景的应用。

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Linearization of the kernel functions in SVM

点击打开题目链接

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 155 Accepted Submission(s): 90

Problem Description

SVM（Support Vector Machine）is an important classification tool, which has a wide range of applications in cluster analysis, community division and so on. SVM The kernel functions used in SVM have many forms. Here we only discuss the function of the form f(x,y,z) = ax^2 + by^2 + cy^2 + dxy + eyz + fzx + gx + hy + iz + j. By introducing new variables p, q, r, u, v, w, the linearization of the function f(x,y,z) is realized by setting the correspondence x^2 <-> p, y^2 <-> q, z^2 <-> r, xy <-> u, yz <-> v, zx <-> w and the function f(x,y,z) = ax^2 + by^2 + cy^2 + dxy + eyz + fzx + gx + hy + iz + j can be written as g(p,q,r,u,v,w,x,y,z) = ap + bq + cr + du + ev + fw + gx + hy + iz + j, which is a linear function with 9 variables.

Now your task is to write a program to change f into g.

Input

The input of the first line is an integer T, which is the number of test data (T<120). Then T data follows. For each data, there are 10 integer numbers on one line, which are the coefficients and constant a, b, c, d, e, f, g, h, i, j of the function f(x,y,z) = ax^2 + by^2 + cy^2 + dxy + eyz + fzx + gx + hy + iz + j.

Output

For each input function, print its correspondent linear function with 9 variables in conventional way on one line.

Sample Input

  
   2
0 46 3 4 -5 -22 -8 -32 24 27
2 31 -5 0 0 12 0 0 -49 12

Sample Output

  
   46q+3r+4u-5v-22w-8x-32y+24z+27
2p+31q-5r+12w-49z+12

Source

2014上海全国邀请赛——题目重现（感谢上海大学提供题目）

水题一只，不过贡献了好多wa，细节没考虑完全：

做到以下几点应该没问题了：

1、第一个字母前面的系数如果是正数则不需要输出‘+’；

2、如果系数是1或-1，那么对于前九项的系数不输出1只有‘+’或‘-’，常数则正常输出；

3、如果全部为0时应该输出0，不应该什么也不输出；

4、如果前九项的系数为0，最后常数的系数为一个正数，则不需要输出‘+’；

代码：

#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<stdio.h>
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
#define N 9
using namespace std;
char a[N]= {'p','q','r','u','v','w','x','y','z'};
int main()
{
    int t,b[N+1],i;
    bool flag=false;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            scanf("%d%d%d%d%d%d%d%d%d%d",&b[0],&b[1],&b[2],&b[3],&b[4],&b[5],&b[6],&b[7],&b[8],&b[9]);
            flag=false;
            for(i=0; i<N; i++)
            {
                if(!b[i])
                    continue;
                if(flag||b[i]<0)
                    printf("%c",b[i]>0?'+':'-');
                flag=true;
                if(!(b[i]==1||b[i]==-1))
                {
                    printf("%d",b[i]>0?b[i]:-b[i]);
                }
                printf("%c",a[i]);
            }
            if(!b[i]&&!flag)
            {
                printf("0\n");
                continue;
            }
            if(b[i])
            {
                if(flag||b[i]<0)
                printf("%c",b[i]>0?'+':'-');
                printf("%d",b[i]>0?b[i]:-b[i]);
            }
            printf("\n");
        }
    }
    return 0;
}