NC15665 maze
题目地址:
基本思路:
从题目意思可以看出这是一个最短路问题,所以可以建图跑Dijkstra,但是数据并不大而建图比较麻烦,所以我们可以直接用Bfs模拟spfa算法的过程,把Bfs稍微修改其实就是spfa了。
参考代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h> using namespace std; #define IO std::ios::sync_with_stdio(false) #define int long long #define rep(i, l, r) for (int i = l; i <= r; i++) #define per(i, l, r) for (int i = l; i >= r; i--) #define mset(s, _) memset(s, _, sizeof(s)) #define pb push_back #define pii pair <int, int> #define mp(a, b) make_pair(a, b) #define INF 0x3f3f3f3f inline int read() { int x = 0, neg = 1; char op = getchar(); while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); } while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); } return neg * x; } inline void print(int x) { if (x < 0) { putchar('-'); x = -x; } if (x >= 10) print(x / 10); putchar(x % 10 + '0'); } const int maxn = 310; int n,m,q,d[4][2] = {{0,1},{1,0},{0,-1},{-1,0}}; char a[maxn][maxn]; bool vis[maxn][maxn]; int dp[maxn][maxn]; map<pii,vector<pii>> memo; pii star,ed; int bfs(){ mset(dp,INF); mset(vis,false); queue<pii> que; dp[star.first][star.second] = 0; que.push(star); while(!que.empty()){ int x = que.front().first,y = que.front().second; que.pop(); vis[x][y] = false;//vis数组是用来防止重复入队的; if(memo.count({x,y})) { for(auto it : memo[{x,y}]) {//传送门的情况; int nx = it.first, ny = it.second; if (nx >= 1 && nx <= n && ny >= 1 && ny <= m && a[nx][ny] != '#' && !vis[nx][ny]) { if (dp[x][y] + 3 <= dp[nx][ny]) { dp[nx][ny] = min(dp[nx][ny], dp[x][y] + 3); vis[nx][ny] = true; que.push({nx, ny}); } } } } for(int i = 0 ; i < 4 ; i++){//直接走的情况; int nx = x + d[i][0],ny = y + d[i][1]; if(nx >= 1 && nx <= n && ny >= 1 && ny <= m && a[nx][ny] != '#' && !vis[nx][ny]){ if(dp[x][y] + 1 <= dp[nx][ny]){ dp[nx][ny] = min(dp[nx][ny],dp[x][y] + 1); vis[nx][ny] = true; que.push({nx,ny}); } } } } return dp[ed.first][ed.second] >= INF ? -1 : dp[ed.first][ed.second]; } signed main() { IO; while (cin >> n >> m >> q) { rep(i, 1, n) { rep(j, 1, m) { cin >> a[i][j]; if (a[i][j] == 'S') star = {i, j}; if (a[i][j] == 'T') ed = {i, j}; } } memo.clear(); rep(i, 1, q) { int rx, ry, tx, ty; cin >> rx >> ry >> tx >> ty; rx++, ry++, tx++, ty++;//输入的数据是从0开始的要改一下; memo[{rx, ry}].push_back({tx, ty}); } int ans = bfs(); cout << ans << '\n'; } return 0; }