Fire Net

Time Limit : 2000/1000ms (Java/Other)???Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 10???Accepted Submission(s) : 3

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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.. .... 2 XX .X 3 .X. X.X .X. 3 ... .XX .XX 4 .... .... .... .... 0

Sample Output

5 1 5 2 4 題意： 給出你正方形的邊長格數(shù)（n<=4)，并在隨機在某些位置放上墻，除了由墻隔開的，一個十字架方向上不能重，問你最多能放多少架大炮。 分析： 計算出每個空位的十字架方向上有多少個空位，即：將會有多少會與其沖突。然后排序選擇十字架方向上空位最少的位置，（最少的位置干擾最少，最有可能架大炮）開始判斷是否能夠放大炮。 因此，該函數(shù)包括兩個最重要的函數(shù)：count（計數(shù)），bool right（判斷是否正確） 計數(shù)時：要判斷四個方向，遇墻break。 bool right：也要判斷四個方向，遇墻break，如果遇到map[][]=1,說明該圖此位置已有大炮，返回false，否則檢測到最后都行返回true。如果該位置可以的話要將map中的數(shù)置為1，作為標記。 代碼：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;/*
int cmp(const void *a,const void *b){
??????? return *(int*)a-*(int*)b;
}*/
int count(int a[4][4],int row,int col,int n){/*數(shù)數(shù)，計算每個空位，十字架方向上所有能夠放的，即會發(fā)生沖突的數(shù)目*/
??? int count1=0,i;/*注意四個方向，count計數(shù)，別忘了設(shè)初值為0*/
??? for(i=row-1;i>=0;i--){/*up*/
??????? if(a[i][col]==2)/*碰到墻才會停止*/
??????????? break;
??????? else
??????????? count1++;
??? }
??? for(i=row+1;i<n;i++){/*down*/
??????? if(a[i][col]==2)
??????????? break;
??????? else
??????????? count1++;
??? }
??? for(i=col-1;i>=0;i--){/*left*/
??????? if(a[row][i]==2)
??????????? break;
??????? else
??????????? count1++;
??? }
??? for(i=col+1;i<n;i++){/*right*/
??????? if(a[row][i]==2)
??????????? break;
??????? else
??????????? count1++;
??? }
??? return count1;
}
bool right(int a[4][4],int row,int col,int n){/*判斷是否可以安置炮*/
??? int i;
??????? for(i=row-1;i>=0;i--){/*用bool型*/
??????? if(a[i][col]==1)/*十字架方向上碰到架炮的地方，就說明有沖突，即此處不可行*/
??????????? return false;
??????? else if(a[i][col]==2)/*碰到墻停止*/
??????????? break;
??? }
??? //down
??? for(i=row+1;i<n;i++){/*仍舊是四個方向上檢測*/
??????? if(a[i][col]==1)
??????????? return false;
??????? else if(a[i][col]==2)
??????????? break;
??? }
??? //left
??? for(i=col-1;i>=0;i--){
??????? if(a[row][i]==2)
??????????? break;
??????? else if(a[row][i]==1)
??????????? return false;
??? }
??? //right
??? for(i=col+1;i<n;i++){
??????? if(a[row][i]==2)
??????????? break;
??????? else if(a[row][i]==1)
??????????? return false;
??? }
??? return true;
}
int main(){
??? int n,i,j,k,ans;
??? char c;
??? int map[4][4];
??? int cnt[4][4];
??? while(scanf("%d",&n),n){
??????? k=0,ans=0;
??????? memset(cnt,-1,sizeof(cnt));
??????? for(i=0;i<n;i++){
??????????? for(j=0;j<n;j++){
?????????????? // scanf("%c",&c);
??????????????? cin>>c;
??????????? if(c=='.'){/*將字符型的圖轉(zhuǎn)化成數(shù)組型的*/
??????????????? k++;/*順表記錄下總共有多少個空的位置，方便最后檢測成不成功時的循環(huán)次數(shù)*/
??????????????? map[i][j]=0;/*0表示可放的位置*/
??????????? }
??????????? else if(c=='X')/*X表示墻的位置*/
??????????????? map[i][j]=2;
??????????? }
??????? }
??????? for(i=0;i<n;i++){/*計數(shù)，遇墻跳過，遇空地計數(shù)，并將地址與數(shù)目用二維數(shù)組cnt[i][j]保留下來*/
??????????? for(j=0;j<n;j++){
??????????????? if(map[i][j]==2)
??????????????????? continue;
??????????????? else
??????????????????? cnt[i][j]=count(map,i,j,n);
??????????? }
??????? }
??????? /*qsort(cnt,n*n,sizeof(cnt[0][0]),cmp);//！！！不能使用qsort直接對cnt排序，否則將會把位置給弄丟，所以應(yīng)該使用冒泡排序*/
???????? int min=7,mini,minj;/*最多只能是6，用題中的最大可能做最小箱的箱底*/
???????? while(k--){/*有k個空位，即有k種可能性，每個位置都要判斷下*/
??????????? for(i=0;i<n;i++){
??????????????? for(j=0;j<n;j++){
??????????????????? if(cnt[i][j]==-1)
??????????????????????? continue;
??????????????????? else if(cnt[i][j]<min){/*要先排排序，從最小的開始，使用貪心，因為周圍的可能最少，該位成功的可能性最大*/
??????????????????????? min=cnt[i][j];/*重置最小*/
??????????????????????? mini=i;/*記錄位置*/
??????????????????????? minj=j;
??????????????????? }
??????????????? }
??????????? }
??????????? if(right(map,mini,minj,n)){/*判斷是否能夠放*/
??????????????? map[mini][minj]=1;/*可以的話，就將map中的該位置為1，標志為有炮，便于剩下位置的判斷*/
??????????????? ans++;/*成立計數(shù)加1*/
??????????? }
??????????? cnt[mini][minj]=-1;/*該位置不用再檢測了，在cnt數(shù)組中標記下-1，直接跳過，檢測第二個最小的位置*/
??????????? min=7;/*重置最小*/

??????? }
??????? /*“。”點表示可放東西，總計有多少個空，k個*/
/*??????????? for(i=0;i<n;i++){
??????????????? for(j=0;j<n;j++){
??????????????????? if(cnt[i][j]==-1)
??????????????????????? continue;
??????????????????????? else if(right(map,i,j,n)){
??????????????????????????? map[i][j]=1;
??????????????????????????? ans++;
??????????????????????? }

??????????????? }
??????????? }*/
??????? printf("%d\n",ans);/*輸出*/
??? }

return 0;
}

轉(zhuǎn)載于:https://www.cnblogs.com/money-lady/p/3659076.html

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