poj 1753 Flip Game

Flip Game

題意:和poj 1681同樣，4*4的01矩陣輸入爲b|w,關聯也是周圍四個方向。只是最優解(操做個數最少)是隻要最終同色就行，沒有肯定哪一個是1.

此題確實能夠直接枚舉第一行，

思路:本來認爲弄成自由變元的個數以後，所有變爲0和所有變爲1的操做數之和就是var - ret(自由變元的個數)，這樣只須要調用Gauss一次便可。可是還有一個前提就是認爲最有解就是自由變元就認爲是沒操做，可是這一題與1681Painter's Problem 不一樣，必定要枚舉自由變元來肯定其餘維度的變量。因此仍是要兩次調用Gauss().

(Gauss-Jordan elimination)：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define inf 0x3f3f3f3f
int dir[2][4] = {{0,1,0,-1},{1,0,-1,0}};
int a[20][20];
int equ,var;
int x[20],free_var[20];
void debug()
{
    puts("********");
    int i,j;
    rep0(i,0,equ){
        rep1(j,0,var)
            cout<<a[i][j]<<" ";
        cout<<endl;
    }puts("********");
}
int Gauss()
{
    int i,j,k,row,col,cnt = 0;
    for(row = 0,col = 0;row < equ && col < var;row++,col++){
        int mx = row;
        rep0(j,row+1,equ)
            if(abs(a[j][col]) > abs(a[mx][col]))  mx = j;
        if(a[mx][col] == 0){
            row--;  // 行不變；不能經過這裏記錄自由變元的個數，只能記錄沒用的col
            free_var[cnt++] = col;//記錄自由變元的標號；
            continue;
        }
        if(mx != row)
            rep1(k,col,var)
                swap(a[row][k],a[mx][k]);
        rep0(j,row+1,equ){
            if(a[j][col]){
                rep1(k,col,var)
                    a[j][k] ^= a[row][k];
            }
        }
    }
    //debug();
    rep0(i,row,equ)
        if(a[i][var] != 0) return -1;    //無解
    //枚舉自由變元,row表示有用的方程數方程,可是要在判斷出有解的前提下才能說有多組解；
    //if(row < var) return var - row;   //當不須要枚舉時，直接返回自由變元的個數
    int ans = inf,tot = 1 <<(var - row);
    rep0(i,0,tot){
        int cnt = 0,tmp = i;
        rep0(j,0,var - row){
            x[free_var[j]] = (tmp&1);
            if(x[free_var[j]]) cnt++;//**
            tmp >>= 1;
        }
        rep_1(i,row-1,0){
            x[i] = a[i][var];//如今賦爲a[i][var],若爲自由變元以後仍是會等於0，不會重複計算；
            rep0(j,i+1,equ){
                x[i] ^= (a[i][j] && x[j]);  //第j個燈會影響到第i盞燈，同時第j盞燈也會亮
            }
            if(x[i]) cnt++;
        }
        ans = min(ans,cnt);
    }
    return ans;
}
void init(int n)
{
    MS0(x);MS0(a);MS0(free_var);
    int i,j,k;
    rep0(i,0,n)
        rep0(j,0,n){
            int id = i*n+j;
            a[id][id] = 1;
            rep0(k,0,4){
                int nx = i + dir[0][k] ,ny = j + dir[1][k];
                if(nx < 0 || nx >= n || ny < 0 || ny >= n) continue;
                a[nx*n+ny][id] = 1;
            }
        }
}
int tmp[20];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("data2.txt","w",stdout);
    int n = 4,i,id = 0;
    equ = var = n*n;
    rep0(i,0,var){
        char c = getchar();
        if(c == 'w') id |= (1<<i);//使用狀壓來存儲狀態。
        else if(c != 'b')  i--;
    }
    init(n);
    rep0(i,0,var){
         a[i][var] = (id >> i)& 1;
    }
    //debug();
    int ans = inf;
    rep0(j,0,2){
        int ret = Gauss();
        if(ret == -1) continue;        
        ans = min(ans,ret);
        if(j == 2) break;
        init(n);
        rep0(i,0,var){
            a[i][var] = (id & 1)^1;
            id >>= 1;
        }
    }
    if(ans == inf) puts("Impossible");
    else printf("%d\n",ans);
    return 0;
}

View Code

dfs枚舉1:枚舉所有的格子狀態，即1<<16範圍；750ms..

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
int dir[2][4] = {{0,1,0,-1},{1,0,-1,0}};
#define inf 0x3f3f3f3f
int a[5][5],b[5][5];
int check(int s,int id)
{
    int cnt = 0;
    rep0(i,0,4) rep0(j,0,4)   b[i][j] = a[i][j];
    rep0(i,0,16){
        if(s&(1<<i)){
            cnt++;
            int x = i/4,y = i % 4;
            b[x][y] ^= 1;
            rep0(j,0,4){
                int nx = x + dir[0][j],ny = y + dir[1][j];
                if(nx >= 0 && nx < 4 && ny >= 0 && ny < 4)
                    b[nx][ny] ^= 1;
            }
        }
    }
    rep0(i,0,4) rep0(j,0,4) if(b[i][j] != id ) return inf;
    return cnt;
}
int solve(int id)
{
    int i,j,tot = 1<<16,ans = inf;
    rep0(i,0,tot){
        ans = min(ans,check(i,id));
    }
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("data2.txt","w",stdout);
    int n = 4;
    char c[5];
    rep0(i,0,4){
        scanf("%s",c);
        rep0(j,0,4)
            if(c[j] == 'w') a[i][j] = 1;
            else a[i][j] = 0;
    }
    int ans = inf;
    ans = min(ans,solve(0));
    ans = min(ans,solve(1));
    if(ans == inf) puts("Impossible");
    else printf("%d\n",ans);
    return 0;
}

View Code

dfs枚舉第一行:坑爹的WA了..要是看出bug，請評論..

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
int dir[2][4] = {{0,1,0,-1},{1,0,-1,0}};
#define inf 0x3f3f3f3f
int a[5][5],b[5][5];
int check(int s,int id)
{
    int cnt = 0;
    rep0(i,0,4) rep0(j,0,4)   b[i][j] = a[i][j];
    rep0(i,0,4){
        if(s&(1<<i)){
            cnt++;
            b[0][i] ^= 1;
            rep0(j,0,4){
                int nx = 0 + dir[0][j],ny = i + dir[1][j];
                if(nx >= 0 && nx < 4 && ny >= 0 && ny < 4)
                    b[nx][ny] ^= 1;
            }
        }
    }
    rep0(i,1,3){
        rep0(j,0,4)if(b[i-1][j] != id){
            cnt++;
            b[i][j] ^= 1;
            rep0(k,0,4){
                int nx = i + dir[0][k],ny = j + dir[1][k];
                if(nx >= 0 && nx < 4 && ny >= 0 && ny < 4)
                    b[nx][ny] ^= 1;
            }
        }
    }
    rep0(i,0,4) rep0(j,0,4) if(b[i][j] != id) return inf;
    return cnt;
}
int solve(int id)
{
    int tot = 1<<4,ans = inf;
    rep0(i,0,tot){
        ans = min(ans,check(i,id));
    }
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("data2.txt","w",stdout);
    char c[5];
    rep0(i,0,4){
        scanf("%s",c);
        rep0(j,0,4)
            if(c[j] == 'w') a[i][j] = 1;
            else a[i][j] = 0;
    }
    int ans = inf;
    ans = min(ans,solve(0));
    ans = min(ans,solve(1));
    if(ans == inf) puts("Impossible");
    else printf("%d\n",ans);
    return 0;
}

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