HDU1045 贪心+搜索

Fire Net

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9427    Accepted Submission(s): 5493

Problem Description

Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall.

A blockhouse is a small castle that has four openings through which to shoot. The four openings are facing North, East, South, and West, respectively. There will be one machine gun shooting through each opening.

Here we assume that a bullet is so powerful that it can run across any distance and destroy a blockhouse on its way. On the other hand, a wall is so strongly built that can stop the bullets.

The goal is to place as many blockhouses in a city as possible so that no two can destroy each other. A configuration of blockhouses is legal provided that no two blockhouses are on the same horizontal row or vertical column in a map unless there is at least one wall separating them. In this problem we will consider small square cities (at most 4x4) that contain walls through which bullets cannot run through.

The following image shows five pictures of the same board. The first picture is the empty board, the second and third pictures show legal configurations, and the fourth and fifth pictures show illegal configurations. For this board, the maximum number of blockhouses in a legal configuration is 5; the second picture shows one way to do it, but there are several other ways.



Your task is to write a program that, given a description of a map, calculates the maximum number of blockhouses that can be placed in the city in a legal configuration.

 

Input

The input file contains one or more map descriptions, followed by a line containing the number 0 that signals the end of the file. Each map description begins with a line containing a positive integer n that is the size of the city; n will be at most 4. The next n lines each describe one row of the map, with a '.' indicating an open space and an uppercase 'X' indicating a wall. There are no spaces in the input file.

 

Output

For each test case, output one line containing the maximum number of blockhouses that can be placed in the city in a legal configuration.

 

Sample Input

4.X......XX......2XX.X3.X.X.X.X.3....XX.XX4................0

 

Sample Output

51524

可以用深搜做，有点类似回溯，把每个点作为第一个搜索的起点来搜，搜完了再回溯，不过记得要把搜完的点复原，因为在下一个点来看上一个点是没动过的。

#include<stdio.h>//城市的尺寸 int n;//城市的地图，最多是4*4char map[4][4];//最多放的炮塔数int bestn;//看炮塔是否能够放置 int canput(int row, int col){int i;for (i = row - 1; i >= 0; i--){if (map[i][col] == 'X'){break;}if (map[i][col] == 'o'){return 0;}}for (i = col - 1; i >= 0; i--){if (map[row][i] == 'X'){break;}if (map[row][i] == 'o'){return 0;}}return 1;}//K表示放置炮塔的位置 void backtrack(int k, int current){int x, y;if (k >= n*n){if (current>bestn){bestn = current;}return;}else{x = k / n;y = k%n;if (map[x][y] == '.'&&canput(x, y)){map[x][y] = 'o';backtrack(k + 1, current + 1);map[x][y] = '.';}backtrack(k + 1, current);}}void initial(){int i, j;for (i = 0; i<4; i++){for (j = 0; j<4; j++){map[i][j] = '.';}}}int main(){scanf("%d", &n);while (n){int i, j;bestn = 0;initial();for (i = 0; i<n; i++){for (j = 0; j<n; j++){char ch;ch = getchar();if (ch == '\n'){j--;continue;}else{map[i][j] = ch;}}}backtrack(0, 0);printf("%d\n", bestn);scanf("%d", &n);}return 0;}

除此之外还有一个巧妙的做法，用二分图匹配来做，这里给出用匈牙利算法求得的结果。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int MAXN = 105;const int oo = 1e9;bool G[MAXN][MAXN], used[MAXN];int p[MAXN], x, y;struct node{int x, y;}a[MAXN][MAXN];bool Find(int u){///匈牙利算法匹配    for(int i=1; i<=y; i++)    {        if(G[u][i] && used[i] == false)        {            used[i] = true;            if(!p[i] || Find(p[i]))            {                p[i] = u;                return true;            }        }    }    return false;}int main(){    int N;    while(scanf("%d", &N), N)    {        int i, j;        char s[MAXN][MAXN];        for(i=0; i<N; i++)            scanf("%s", s[i]);        x = y = 0;        for(i=0; i<N; i++)        for(j=0; j<N; j++)        {///把图分割，以相连的‘.’为行和列重新分配编号            if(s[i][j] == '.')            {                if(j == 0 || s[i][j-1] == 'X')                    x++;                a[i][j].x = x;            }            if(s[j][i] == '.')            {                if(j == 0 || s[j-1][i] == 'X')                    y++;                a[j][i].y = y;            }        }        memset(G, 0, sizeof(G));        for(i=0; i<N; i++)        for(j=0; j<N; j++)        {            if(s[i][j] == '.')            {                int u = a[i][j].x;                int v = a[i][j].y;                G[u][v] = true;///用行匹配列            }        }        int ans = 0;        memset(p, 0, sizeof(p));        for(i=1; i<=x; i++)        {            memset(used, false, sizeof(used));            if( Find(i) == true )                ans++;        }        printf("%d\n", ans);    }    return 0;}