Codeforces 677D Vanya and Treasure 暴力+BFS

链接

Codeforces 677D Vanya and Treasure

题意

n*m中有p个type，经过了任意一个 type=i 的各自才能打开 type=i+1 的钥匙，最初有type=1的钥匙, 问拿到type=p的钥匙最少需要走多少步

思路

第一想法就是按type来递推, 将type相同的存到一起，dp[i][j]=min(dp[i][j], dp[k][l]+distance([i][j], [k][l]))，其中 a[i][j] = a[k][l]+1. 但这样type相同的个数较多时就超时了，所以根据type的个数，较多时使用BFS就可以了。 我没有仔细研究时间复杂度就过了，官方题解好像有均摊复杂度的证明。

代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
#include <cstring>
#include <string>

#define LL long long
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MOD 1000000007
using namespace std;

struct POS{
	int x, y;
	POS(int x = 0, int y = 0) :x(x), y(y){};
};
vector<POS> s[90005];
int a[305][305];
int d[305][305];
bool vis[305][305];
int dp[305][305];
const int step[4][2] = { 1, 0, 0, 1, -1, 0, 0, -1 };
int get_distance(POS &a, POS &b){
	return abs(a.x - b.x) + abs(a.y - b.y);
}
int main(){
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
	int n, m, p;
	while (~scanf("%d%d%d", &n, &m, &p)){
		int x;
		for (int i = 1; i <= n; ++i){
			for (int j = 1; j <= m; ++j){
				scanf("%d", &a[i][j]);
				s[a[i][j]].push_back(POS(i, j));
			}
		}
		memset(dp, INF, sizeof(dp));
		for (int i = 0; i < s[1].size(); ++i){
			dp[s[1][i].x][s[1][i].y] = s[1][i].x + s[1][i].y - 2;
		}
		for (int i = 2; i <= p; ++i){
			if (s[i].size() * s[i - 1].size() < n * m){
				for (int j = 0; j < s[i].size(); ++j){
					POS J = s[i][j];
					for (int k = 0; k < s[i - 1].size(); ++k){
						POS K = s[i - 1][k];
						dp[J.x][J.y] = min(dp[K.x][K.y] + get_distance(J, K), dp[J.x][J.y]);
					}
				}
			}
			else{
				memset(d, INF, sizeof(d));
				memset(vis, 0, sizeof(vis));
				queue<POS> Q;
				for (int j = 0; j < s[i - 1].size(); ++j){
					POS J = s[i - 1][j];
					Q.push(J);
					vis[J.x][J.y] = true;
					d[J.x][J.y] = dp[J.x][J.y];
				}
				while (!Q.empty()){
					POS u = Q.front();
					Q.pop();
					vis[u.x][u.y] = false;
					for (int j = 0; j < 4; ++j){
						int x = u.x + step[j][0];
						int y = u.y + step[j][1];
						if (a[u.x][u.y] == i) dp[u.x][u.y] = min(dp[u.x][u.y], d[u.x][u.y]);
						if (x < 1 || x >  n || y < 1 || y > m) continue;
						if (d[x][y] > d[u.x][u.y] + 1){
							d[x][y] = d[u.x][u.y] + 1;
							if (!vis[x][y]){
								vis[x][y] = true;
								if (a[x][y] == i){
									dp[x][y] = min(dp[x][y], d[x][y]);
								}
								Q.push(POS(x, y));
							}
						}
					}
				}
			}
		}
		POS K = s[p][0];
		cout << dp[K.x][K.y] << endl;
	}
}