UVa 11953 - Battleships

来源：互联网发布：韩国身份证汉字知乎编辑：程序博客网时间：2024/06/05 05:59

題目：打飛機，打船（中學時經常玩的）：兩個人在紙上畫上船，分別盲報坐標打對方的船。

已知船的被擊中的點和未必機中的點，判斷存活的船的數量。

分析：圖論，搜索。直接利用dfs搜索即可。

說明：好久沒有刷題了╮(╯▽╰)╭。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>using namespace std;char grid[101][101];int dxy[4][2] = { 1, 0, 0, 1, -1, 0, 0, -1};int dfs(int x, int y, int n){if (x < 0 || x >= n || y < 0 || y >= n) {return 0;}if (grid[x][y] == '.' || grid[x][y] == '#') {return 0;}int flag = 0;if (grid[x][y] == 'x') {flag = 1;}grid[x][y] = '#';for (int i = 0; i < 4; ++ i) {if (dfs(x+dxy[i][0], y+dxy[i][1], n)) {flag = 1;}}return flag;}int main(){int T, N;while (~scanf("%d",&T)) {for (int t = 1; t <= T; ++ t) {scanf("%d",&N);for (int i = 0; i < N; ++ i) {scanf("%s",grid[i]);}int count = 0;for (int i = 0; i < N; ++ i) {for (int j = 0; j < N; ++ j) {if (grid[i][j] == 'x' || grid[i][j] == '@') {count += dfs(i, j, N);}}}printf("Case %d: %d\n",t,count);}}return 0;}

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