HDU 4198 Quick out of the Harbour(优先队列 + bfs）

解题思路：

直接的bfs，因为和时间有关，需要采用优先队列.

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <stack>
using namespace std;
const int MAXN = 500 + 10;
char G[MAXN][MAXN];
int N, M, D;
struct Node
{
	int x, y, time;
	bool operator < (const Node& rhs)const
	{
		return time > rhs.time;
	}
};
bool vis[MAXN][MAXN];
int g[][2] = {{-1,0},{1,0},{0,-1},{0,1}};
int bfs(int sx, int sy)
{
	memset(vis, false, sizeof(vis));
	priority_queue<Node> Q;
	while(!Q.empty()) Q.pop();
	vis[sx][sy] = 1;
	Node u, v;
	u.x = sx, u.y = sy, u.time = 1;
	Q.push(u);
	while(!Q.empty())
	{
		u = Q.top();
		Q.pop();
		for(int i=0;i<4;i++)
		{
			int x = u.x + g[i][0];
			int y = u.y + g[i][1];
			if(x < 1 || x > N || y < 1 || y > M)
				return u.time;
			if(vis[x][y] || G[x][y] == '#')
				continue;
			if(G[x][y] == '@')
			{
				v.x = x, v.y = y, v.time = u.time + D + 1;
				vis[x][y] = true;
				Q.push(v);
			}
			if(G[x][y] == '.')
			{
				vis[x][y] = true;
				v.x = x, v.y = y, v.time = u.time + 1;
				Q.push(v);
			}
		}
	}
	return -1;
}

int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		int sx, sy;
		scanf("%d%d%d", &N, &M, &D);
		for(int i=1;i<=N;i++)
		{
			for(int j=1;j<=M;j++)
			{
				scanf(" %c", &G[i][j]);
				if(G[i][j] == 'S')
				{
					sx = i;
					sy = j;
				}
			}
		}
		/*for(int i=1;i<=N;i++)
		{
			for(int j=1;j<=M;j++)
				cout<<G[i][j];
			cout<<endl;
		}*/
		int ans = bfs(sx, sy);
	 	printf("%d\n", ans);
	}
	return 0;
}