poj 3185 The Water Bowls(高斯消元)

http://poj.org/problem?id=3185

输入变成一维的了，转化成 n*(n+1)的系数矩阵。

题目中说方案数一定存在。枚举自由元求最优解即可。

#include <stdio.h>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#define LL long long
#define _LL __int64

using namespace std;
const int INF = 0x3f3f3f3f;

int a[22][22];
int equ,var;
int free_num;
int free_x[22];
int x[22];
int Gauss()
{
    int row,col,i,j,max_r;
    row = col = 0;
    free_num = 0;
    while(row < equ && col < var)
    {
        max_r = row;
        for(i = row+1; i < equ; i++)
            if(abs(a[i][col]) > abs(a[max_r][col]))
                max_r = i;
        if(max_r != row)
        {
            for(j = col; j < var+1; j++)
                swap(a[max_r][j],a[row][j]);
        }
        if(a[row][col] == 0)
        {
            free_x[ free_num++ ] = col;
            col++;
            continue;
        }

        for(i = row+1; i < equ; i++)
        {
            if(a[i][col] == 0) continue;
            for(j = col; j < var+1; j++)
                a[i][j] ^= a[row][j];
        }
        row++;
        col++;
    }

    for(i = row; i < equ; i++)
        if(a[i][col] != 0)
            return -1;

    if(row < var)
        return var-row;

    for(i = var-1; i >= 0; i--)
    {
        x[i] = a[i][var];
        for(j = i+1; j < var; j++)
            x[i] ^= (a[i][j] && x[j]);
    }
    return 0;
}

void solve()
{
    int t = Gauss();
    int ans;

    if(t == 0)
    {
        ans = 0;
        for(int i = 0; i < var; i++)
            ans += x[i];
        printf("%d\n",ans);
        return;
    }

    else if(t > 0)
    {
        ans = INF;
        int sta = (1<<t);

        for(int i = 0; i < sta; i++)
        {
            int cnt = 0;
            for(int j = 0; j < t; j++)
            {
                if( (1<<j) & i)
                {
                    x[ free_x[j] ] = 1;
                    cnt++;
                }
                else
                    x[ free_x[j] ] = 0;
            }
            int k,l;
            for(int j = var-1-t; j >= 0; j--)
            {
                for(k = j; k < var; k++)
                    if(a[j][k]) break;

                x[k] = a[j][var];
                for(l = k+1; l < var; l++)
                    x[k] ^= (a[j][l] && x[l]);
                cnt += x[k];
            }
            ans = min(ans,cnt);
        }
        printf("%d\n",ans);
        return;
    }

}

int main()
{
    memset(a,0,sizeof(a));
    memset(x,0,sizeof(x));
    for(int i = 0; i < 20; i++)
        scanf("%d",&a[i][20]);
    equ = var = 20;

    for(int i = 0; i < 20; i++)
    {
        a[i][i] = 1;
        if(i > 0)
            a[i-1][i] = 1;
        if(i < 19)
            a[i+1][i] = 1;
    }
    solve();
    return 0;
}

poj 3185 The Water Bowls(高斯消元),布布扣,bubuko.com

poj 3185 The Water Bowls(高斯消元)

原文：http://blog.csdn.net/u013081425/article/details/24556687