blockhouses
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blockhouses
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blockhouses
時間限制:1000?ms ?|? 內存限制:65535?KB 難度:3 描述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.? 輸入
2012年10月份月賽(高年級組)
x代表墻,其余的地方建造炮彈,每一行,每一列只能有一個,除非他們中間有墻
AC代碼:
#include <bits/stdc++.h> using namespace std; char mp[5][5]; int n,i,j; int maxn; int flag; int cha(int x,int y){flag=1;for(i=x;i>=0;i--){ //查詢同列的上邊是否有blockhouses if(mp[i][y]=='X')break;if(mp[i][y]=='b'){flag=0;break;}}for(i=y;i>=0;i--){ //查詢同行的左邊是否有blockhousesif(mp[x][i]=='X')break;if(mp[x][i]=='b'){flag=0;break;}}return flag; } int DFS(int a,int b){int i=a/n,j=a%n;if(a==n*n){ //控制搜索結束 maxn=max(maxn,b);return maxn;}if(mp[i][j]=='.'&&cha(i,j)==1){mp[i][j]='b';DFS(a+1,b+1);mp[i][j]='.';}DFS(a+1,b); } int main() {while(scanf("%d",&n),n){memset(mp,0,sizeof(mp));for(i=0;i<n;i++)scanf("%s",mp[i]);maxn=0;printf("%d\n",DFS(0,0));}return 0; }總結
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