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;}
每一件事都要用多方面的角度来看它
