《挑战程序设计》 P34
第一次使用pair
1.头文件:<utility>
2.成员:mypair.first, mypair.second
3.运算符:<、>、<=、>=、==、!=,其规则是先比较first,first相等时再比较second
4.相关函数:make_pair 例如:p=make_pair(1,1); q.push(make_pair(1,1));
#include <iostream> #include <utility> #include <cstdio> #include <queue> using namespace std; const int INF = 100000000; typedef pair<int, int> P; const int N = 100; char maze[N][N + 1]; int n, m; int sx, sy; int gx, gy; int d[N][N]; //记录到各个位置的最短距离 int dx[4] = {-1, 0, 1, 0}; int dy[4] = {0, 1, 0, -1}; int bfs() { queue<P> que; for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j) d[i][j] = INF; que.push(P(sx, sy)); d[sx][sy] = 0; while (que.size()) { P p = que.front(); que.pop(); if (p.first == gx && p.second == gy) break; for (int i = 0; i < 4; ++i) { int nx = p.first + dx[i]; int ny = p.second + dy[i]; if (nx < n && nx >= 0 && ny < m && ny >= 0 && maze[nx][ny] != '#' && d[nx][ny] == INF) { d[nx][ny] = d[p.first][p.second] + 1; que.push(P(nx, ny)); } } } return d[gx][gy]; } void solve() { scanf("%d%d", &n, &m); for (int i = 0; i < n; ++i) { getchar(); for (int j = 0; j < m; ++j) { scanf("%c", &maze[i][j]); if (maze[i][j] == 'S') sx = i, sy = j; if (maze[i][j] == 'G') gx = i, gy = j; } } int res = bfs(); printf("%d\n", res); } int main() { solve(); return 0; } /**************** Input: 10 10 #S######.# ......#..# .#.##.##.# .#........ ##.##.#### ....#....# .#######.# ....#..... .####.###. ....#...G# Output: 22 *****************/