2.24DFS和BFS刷题

洛谷P2895：用BFS走出危险区域，危险区域存在时间，我们用ma记录最快变成危险区域的时间， 然后每次枚举时间+1然后跟ma数组比较看能不能走，然后时间复杂度为O(305^2)。

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int N = 50010;
int m, x, y, t;
typedef pair<int, int>PII;
int ma[305][305], ans[305][305];
bool v[305][305];
int dx[5] = {1, -1, 0, 0, 0},dy[5] = {0, 0, 1, -1, 0};
int change(int x){
	if(x == -1)return 99999;
	return x;
}
int main(){
	cin >> m;
	memset(ma, -1, sizeof(ma));
	while(m--){
		cin >> x >> y >> t;
		for(int i = 0; i < 5; i++){
			if(x + dx[i] >= 0 && y + dy[i] >= 0 && (ma[x + dx[i]][y + dy[i]] == -1 || ma[x + dx[i]][y + dy[i]] > t)){
				ma[x + dx[i]][y + dy[i]] = t;
			}
		}
	}
	queue<PII> q;
	q.push({0, 0});
	v[0][0] = true;
	while(q.size()){
		auto t = q.front();
		q.pop();
		int xd = t.first, yd = t.second;
		int s = ans[xd][yd] + 1;
		if(ma[xd][yd] == -1){
			cout << s - 1 << endl;
			return 0;
		}
		for(int i = 0; i < 4; i++){
			int xx = xd + dx[i], yy = yd + dy[i];
			if(xx >= 0 && yy >= 0 && s < change(ma[xx][yy]) && !v[xx][yy]){
				v[xx][yy] = true;
				ans[xx][yy] = s;
				q.push({xx, yy});
			}
		}
	}
	cout << -1 << endl;
	
	return 0;
}

洛谷P2036

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int n, ss = 1, bb;
int s[15], b[15];
bool flag[15];
int ans = 0x7ffffff;
void dfs(int x){
	if(x > n){
		
	}
	else{
		for(int i = 1; i <= n; i++){
			if(flag[i] == false){
				ss *= s[i], bb += b[i];
				flag[i] = true;
				ans = min(ans, abs(ss - bb));
				dfs(x + 1);
				ss /= s[i], bb -= b[i];
				flag[i] = false;
			}
		}
	}
}
int main(){
	cin >> n;
	for(int i = 1; i <= n; i++){
		cin >> s[i] >> b[i];
	}
	dfs(1);
	cout << ans << endl;
	
	return 0;
}

洛谷P1605迷宫

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
int n, m, t;
const int N = 15;
int sx, sy, fx, fy, xd, yd;
bool za[N][N], st[N][N];
int ans;
int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};
void dfs(int x, int y){
	if(x == fx && y == fy){
		ans++;
		return;
	}
	for(int i = 0; i < 4; i++){
		int xx = x + dx[i], yy = y + dy[i];
		if(xx >= 1 && xx <= n && yy >= 1 && yy <= m && !za[xx][yy] && !st[xx][yy]){
			st[xx][yy] = true;
			dfs(xx, yy);
			st[xx][yy] = false;
		}
	}
}
int main(){
	cin >> n >> m >> t;
	cin >> sx >> sy >> fx >> fy;
	st[sx][sy] = true;//注意特殊点 
	while(t--){
		cin >> xd >> yd;
		za[xd][yd] = true;
	}
	dfs(sx, sy);
	cout << ans << endl;
	
	return 0;
}