POJ 3185 The Water Bowls

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高斯消元+枚舉自由變元

The Water Bowls Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4276   Accepted: 1678 Description The cows have a line of 20 water bowls from which they drink. The bowls can be either right-side-up (properly oriented to serve refreshing cool water) or upside-down (a position which holds no water). They want all 20 water bowls to be right-side-up and thus use their wide snouts to flip bowls.  Their snouts, though, are so wide that they flip not only one bowl but also the bowls on either side of that bowl (a total of three or -- in the case of either end bowl -- two bowls).  Given the initial state of the bowls (1=undrinkable, 0=drinkable -- it even looks like a bowl), what is the minimum number of bowl flips necessary to turn all the bowls right-side-up? Input Line 1: A single line with 20 space-separated integers Output Line 1: The minimum number of bowl flips necessary to flip all the bowls right-side-up (i.e., to 0). For the inputs given, it will always be possible to find some combination of flips that will manipulate the bowls to 20 0‘s. Sample Input 0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 Sample Output 3 Hint Explanation of the sample:  Flip bowls 4, 9, and 11 to make them all drinkable:  0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [initial state]  0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 4]  0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 [after flipping bowl 9]  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [after flipping bowl 11] Source USACO 2006 January Bronze

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#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int maxn=30;int a[maxn][maxn],x[maxn],ta[maxn];int val0,mim;void dfs(int v){    if(v==20)    {        for(int i=0;i<20;i++)            x[i]=ta[i];        for(int i=val0-1;i>=0;i--)        {            x[i]=a[i][20];            for(int j=i+1;j<20;j++)                x[i]^=(a[i][j]&&x[j]);        }        int temp=0;        for(int i=0;i<20;i++)            if(x[i]) temp++;        mim=min(mim,temp);        return ;    }    ta[v]=0;    dfs(v+1);    ta[v]=1;    dfs(v+1);}void Guass(){    int i,j;    for(i=0,j=0;i<20&&j<20;j++)    {        int k=i;        for(;k<20;k++)            if(a[k][j]) break;        if(a[k][j])        {            if(k!=i)            {                for(int r=0;r<=20;r++)                    swap(a[k][r],a[i][r]);            }            for(int r=0;r<20;r++)            {                if(r!=i&&a[r][j])                {                    for(int c=j;c<=20;c++)                        a[r][c]^=a[i][c];                }            }            i++;        }    }    if(i==20)///無自由變元    {        int ans=0;        for(int k=0;k<20;k++)            if(a[k][20]) ans++;        printf("%d\n",ans);    }    else    {        val0=i; mim=100;        dfs(val0);        printf("%d\n",mim);    }}int main(){    for(int i=0;i<20;i++)    {        int x;        scanf("%d",&x);        a[i][20]=x;        a[i][i]=1;        if(i-1>=0) a[i-1][i]=1;        if(i+1<20) a[i+1][i]=1;    }    Guass();    return 0;}