Description
Input
Output
Sample Input
2 6 5 7 ##### #S..# #@#.# #...# #@### #.### 4 5 3 ##### #S#.# #@..# ###@#
Sample Output
16 11
Source
题意：求走出地图的最短时间，‘#‘不能走,‘.‘耗时一，‘@‘耗时K+1
思路：在BFS的基础上加上优先队列
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int MAXN = 510;
struct Point {
int x, y, step;
bool operator< (Point const &a) const {
return step > a.step;
}
} st, ed;
char map[MAXN][MAXN];
int vis[MAXN][MAXN];
int dx[4]={1, -1, 0, 0};
int dy[4]={0, 0, 1, -1};
int n, m, k;
void bfs() {
memset(vis, 0, sizeof(vis));
priority_queue<Point> q;
vis[st.x][st.y] = 1;
q.push(st);
while (!q.empty()) {
Point cur = q.top();
q.pop();
if (cur.x == 0 || cur.x == n-1 || cur.y == 0 || cur.y == m-1) {
printf("%d\n", cur.step+1);
return;
}
for (int i = 0; i < 4; i++) {
int nx = cur.x + dx[i];
int ny = cur.y + dy[i];
if (vis[nx][ny] || map[nx][ny] == '#')
continue;
if (nx >= 0 && nx < n && ny >= 0 && ny < m) {
vis[nx][ny] = 1;
Point tmp;
if (map[nx][ny] == '.') {
tmp.x = nx, tmp.y = ny;
tmp.step = cur.step + 1;
}
else if (map[nx][ny] == '@') {
tmp.x = nx, tmp.y = ny;
tmp.step = cur.step + k + 1;
}
q.push(tmp);
}
}
}
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%d%d%d", &n, &m, &k);
for (int i = 0; i < n; i++) {
scanf("%s", map[i]);
for (int j = 0; j < m; j++) {
if (map[i][j] == 'S') {
map[i][j] = '.';
st.x = i, st.y = j, st.step = 0;
}
}
}
bfs();
}
return 0;
}HDU - 4198 Quick out of the Harbour (BFS+优先队列),布布扣,bubuko.com
HDU - 4198 Quick out of the Harbour (BFS+优先队列)
原文地址：http://blog.csdn.net/u011345136/article/details/37818095
