位运算+枚举

位运算http://c.biancheng.net/cpp/html/101.html

在很多情况下（比如翻棋子）在搜索时涉及大量枚举往往导致超时，位运算则很好地解决了这个问题，方便又快捷

HDU  1882   Strange Billboard

http://acm.hdu.edu.cn/showproblem.php?pid=1882

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 813    Accepted Submission(s): 348

Problem Description

The marketing and public-relations department of the Czech Technical University has designed a new reconfigurable mechanical Flip-Flop Bill-Board (FFBB). The billboard is a regular two-dimensional grid of R ×C square tiles made of plastic. Each plastic tile is white on one side and black on the other. The idea of the billboard is that you can create various pictures by flipping individual tiles over. Such billboards will hang above all entrances to the university and will be used to display simple pictures and advertise upcoming academic events.

To change pictures, each billboard is equipped with a ”reconfiguration device”. The device is just an ordinary long wooden stick that is used to tap the tiles. if you tap a tile, it flips over to the other side, i.e., it changes from white to black or vice versa. Do you agree this idea is very clever?

Unfortunately, the billboard makers did not realize one thing. The tiles are very close to each other and their sides touch. Whenever a tile is tapped, it takes all neighboring tiles with it and all of them flip over together. Therefore, if you want to change the color of a tile, all neighboring tiles change their color too. Neighboring tiles are those that touch each other with the whole side. All inner tiles have 4 neighbors, which means 5 tiles are flipped over when tapped. Border tiles have less neighbors, of course.



For example, if you have the billboard configuration shown in the left picture above and tap the tile marked with the cross, you will get the picture on the right. As you can see, the billboard reconfiguration is not so easy under these conditions. Your task is to find the fastest way to ”clear” the billboard, i.e., to flip all tiles to their white side.

 

Input

The input consists of several billboard descriptions. Each description begins with a line containing two integer numbers R and C (1 ≤ R, C ≤ 16) specifying the billboard size. Then there are R lines, each containing C characters. The characters can be either an uppercase letter “X” (black) or a dot “.” (white). There is one empty line after each map.
The input is terminated by two zeros in place of the board size.

 

Output

For each billboard, print one line containing the sentence “You have to tap T tiles.”, where T is the minimal possible number of taps needed to make all squares white. if the situation cannot be solved, output the string “Damaged billboard.” instead.

 

Sample Input

5 5XX.XXX.X.X.XXX.X.X.XXX.XX5 5.XX.X.......XXX..X.X..X..1 5...XX5 5...X....XX.XX....X.......8 9..XXXXX...X.....X.X..X.X..XX.......XX.X...X.XX..XXX..X.X.....X...XXXXX..0 0

 

Sample Output

You have to tap 5 tiles.Damaged billboard.You have to tap 1 tiles.You have to tap 2 tiles.You have to tap 25 tiles.

这题刚开始按照一般枚举写  很高兴的 TML了  (ノへ￣、)  代码如下

#include<iostream>#include<cstring>using namespace std;int dir[][2] = { -1, 0, 0, -1, 0, 0, 0, 1, 1, 0 };int n, m, map[16][16], flip[16][16];//flip最终翻转情况的数组//int opt[16][16];void init(){    char x;    memset(map, 0, sizeof(map));    for (int i = 0; i < n; i++){        for (int j = 0; j < m; j++){            cin >> x;            if (x == 'X')                map[i][j] = 1;        }    }}

//用来查看当前棋子的颜色int get_colour(int x, int y){    int a = map[x][y], i;    for (i = 0; i < 5; i++){        int x2 = x + dir[i][0], y2 = y + dir[i][1];        if (x2 >= 0 && x2 < n&&y2 >= 0 && y2 < m){            a += flip[x2][y2];        }    }    return a % 2;}int calc(){    int i, j, cnt = 0;    for (i = 1; i < n; i++){        for (j = 0; j < m; j++){            if (get_colour(i - 1, j))                flip[i][j] = 1;        }    }    for (i = 0; i < m; i++){        if (get_colour(n - 1, i))            return -1;    }    for (i = 0; i < n; i++){        for (j = 0; j < m; j++){            cnt += flip[i][j];        }    }    return cnt;}int main(){    while (cin >> n >> m, n || m){        int cnt = -1, i, j, num;        init();        for (i = 0; i < 1 << m; i++){            memset(flip, 0, sizeof(flip));            for (j = 0; j < m; j++){                flip[0][m - j - 1] = i >> j & 1;            }            num = calc();            if (num >= 0 && (cnt<0 || cnt>num)){                cnt = num;                //memcpy(opt, flip, sizeof(flip));            }        }        if (cnt < 0)            cout << "Damaged billboard.\n";        else{            cout << "You have to tap " << cnt << " tiles.\n";            //for (i = 0; i < n; i++){                //for (j = 0; j < m; j++){                    //cout << opt[i][j] << ' ';                //}                //cout << "\r\n";            //}        }    }    return 0;}

后来才发现这样写时间复杂度是  nm2^n虽然最大测试数据为15 但超时还是难免的

最终纠结了好久的位运算，加上代码优化终于过了

ac  代码如下

/******************************************************&       按位与(交集)|       按位或(并集)^       按位异或(翻转棋子)~       取反<<        左移>>        右移S|1<<i  向集合中加入第i个元素 for(int i=0;i<1<<n;s++)       (枚举集合)*******************************************************/#include<iostream>#include<cstring>using namespace std;#define inf 0x7fffffff#define min(a,b) a<b?a:bint n, m, mpt[17];char map[17][17];void init(){    int i, j;    memset(map, 0, sizeof(map));    memset(mpt, 0, sizeof(mpt));    if (n >= m){        for (i = 0; i < n; i++){            cin >> map[i];            for (j = 0; j < m; j++){                if (map[i][j] == 'X')                    mpt[i] |= 1 << j;                //cout << mpt[i] << endl;                //如果输入字符为X,加入对应元素到集合            }        }    }    else{   //当列大于行,转置矩阵存储        for (i = 0; i < n; i++){            cin >> map[i];            for (j = 0; j < m; j++){                if (map[i][j] == 'X')                    mpt[j] |= 1 << i;            }        }        n ^= m ^= n ^= m;    }}int find_set(){    int minn = inf, tmp[17], sign[17], i, j, k, cnt;    for (i = 0; i < 1 << m; i++){        for (j = 0; j < n; j++)            tmp[j] = mpt[j];        for (j = 0; j < n; j++){            //以每一行的上一行为参照翻转,判断最后一行状态即可            sign[j] = j == 0 ? i : tmp[j - 1];            tmp[j] ^= sign[j];            tmp[j] ^= sign[j] << 1 & ((1 << m) - 1);            tmp[j] ^= sign[j] >> 1;            tmp[j + 1] ^= sign[j];            //一次翻转当前行,左右下(左端边界判断!!!!!)        }        if (!tmp[n - 1]){            cnt = 0;            for (j = 0; j < n; j++){                for (k = sign[j]; k>0; k >>= 1){                    if (k & 1)                        cnt++;                }            }            minn = min(minn, cnt);        }    }    return minn;} int main(){    while (cin >> n >> m, m || n){        init();        int minn = find_set();        if (minn == inf)            cout << "Damaged billboard.\n";        else            cout << "You have to tap " << minn << " tiles.\n";    }    return 0;}

同样的位运算在很多地方都可以应用进来，能寄到出其不意的效果

swust oj 781 牛喝水

牛喝水

Time limit(ms): 1000

Memory limit(kb): 65535

Submission: 4

Accepted: 4

Accepted

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?

Description

Line 1: A single line with 20 space-separated integers

Input

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.

Output

1

0 0 1 1 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0

Sample Input

1

3

Sample Output

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]

题目大意：一只牛能把碗翻转过来，(由于嘴太大 ^_^ )同时会把他左右的碗全部翻转,1代表需要翻转的碗，问最少可以几次把碗全部翻转（0表示）

这道题呐，用dfs 也能做 ，以前就是dfs  a掉的，今天做了位运算后想了想这不就是一个行为1，列为20的位位运算吗，（上面的说过行大于列就转置，不然第一行枚举２^２０，，转置了就只有两种可能了）～～

先上dfs的代码 100ms（反正不低于100）

#include<iostream>using namespace std;int bowl[21], step, flag;int range(){    int i;    for (i = 0; i<20; i++)    if (bowl[i] == 1)        return 0;    return 1;}void turn(int i){    bowl[i] = !bowl[i];    if (i>0)        bowl[i - 1] = !bowl[i - 1];    if (i<19)bowl[i + 1] = !bowl[i + 1];}void DFS(int i, int dp){    if (step == dp)    {        flag = range();        return;    }    if (i >= 20 || flag) return;    turn(i);    DFS(i + 1, dp + 1);    turn(i);    DFS(i + 1, dp);}int main(){    int i;    for (i = 0; i < 20; i++)        cin >> bowl[i];    for (step = 0; step<20; step++)    {        flag = 0;        DFS(0, 0);        if (flag) break;    }    cout << step << endl;    return 0;}

用位运算枚举0ms秒过啊~~~

#include<iostream>#include<cstring>using namespace std;#define inf 0x7fffffff#define min(a,b) a<b?a:bint mpt[20], tmp[20], sign[20];int main(){    int i, j, cnt, minn = inf, x;    for (j = 0; j < 20; j++){        cin >> x;        if (x)            mpt[j] |= 1 << 0;    }

//行为1，列为20直接转置    for (i = 0; i < 1 << 1; i++){        for (j = 0; j < 20; j++)            tmp[j] = mpt[j];        for (j = 0; j < 20; j++){            sign[j] = j == 0 ? i : tmp[j - 1];            tmp[j] ^= sign[j];            tmp[j + 1] ^= sign[j];        }        if (!tmp[19]){            cnt = 0;            for (j = 0; j < 20; j++)            if (sign[j] & 1)                cnt++;        }        minn = min(minn, cnt);    }    cout << minn << endl;    return 0;}