模版总结小全

发布于:2024-06-30 ⋅ 阅读:(15) ⋅ 点赞:(0)

BFS

最短步数问题

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;

const int N = 50;
char g[N][N],d[N][N];
int dx[] = {-1,0,1,0};
int dy[] = {0,1,0,-1};
int n,m;

int bfs(int x,int y){
	queue<pair<int,int> > q;
	q.push({x,y});
	memset(d,-1,sizeof(d));
	d[x][y] = 1;
	while(!q.empty()){
		auto t = q.front();
		q.pop(); 
		for(int i = 0; i < 4; i++){
			int sx = t.first+dx[i];
			int sy = t.second+dy[i];
			if(sx >= 0 && sy >= 0 && sx < n && sy < m && d[sx][sy] == -1 && g[sx][sy] == '.'){
				q.push({sx,sy});
				d[sx][sy] = d[t.first][t.second]+1;
			}
		} 
	}
	return d[n-1][m-1];
}

int main(){
	cin >> n >> m;
	for(int i = 0; i < n; i++) cin >> g[i];
	cout << bfs(0,0) << endl;
	return 0;
} 

洪水填充:连通块问题

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;

const int N = 120;
char g[N][N];
bool d[N][N];
int n,m;
int dx[8] = {-1,-1,-1,0,1,1,1,0};
int dy[8] = {-1,0,1,1,1,0,-1,-1};

int bfs(int sx,int sy){
	queue<pair<int,int>> q;
	q.push({sx,sy});
	g[sx][sy] = '.';
	while(!q.empty()){
		auto h = q.front();
		q.pop();
		for(int i = 0; i < 8; i++){
			int x = h.first + dx[i];
			int y = h.second + dy[i];
			if(x >= 0 && x < n && y >= 0 && y < m && g[x][y] == 'W'){
				q.push({x,y});
				g[x][y] = '.';
			} 
		}
	}
}

int main(){
	cin >> n >> m;
	int ans = 0;
	for(int i = 0; i < n; i++) cin >> g[i];
	for(int i = 0; i < n; i++){
		for(int j = 0; j < m; j++){
			if(g[i][j] == 'W'){
				bfs(i,j);
				ans++;
			}
		}
	}
	cout << ans << endl;
	return 0;
}

最短路:输出路径

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;

const int N = 50;
int a[5][5];
bool st[N][N];

typedef pair<int,int> PII; 

PII h[25];
PII Prev[5][5];
int hh,tt;

int dx[4] = {-1,0,1,0};
int dy[4] = {0,1,0,-1};

int bfs(int sx,int sy){
	hh = 0;
	tt = -1;
	h[++tt] = {sx,sy};
	memset(Prev,-1,sizeof Prev);
	while(hh <= tt){
		PII t = h[hh++];
		for(int i = 0; i < 4; i++){
			int x = t.first + dx[i];
			int y = t.second + dy[i];
			if(x >= 0 && x < 5 && y >= 0 && y < 5 && a[x][y] == 0 && st[x][y] == false){
				h[++tt] = {x,y};
				Prev[x][y] = t;
				st[x][y] = 1;
			}
		}
	}
} 


int main(){
	for(int i = 0; i < 5; i++){
		for(int j = 0; j < 5; j++){
			cin >> a[i][j];
		}
	}
	bfs(4,4);
	PII end(0,0);
	while(true){
		cout << "(" << end.first << ", " << end.second << ")" << endl;
		if(end.first == 4 && end.second == 4) break;
		end = Prev[end.first][end.second];
		
	}
}

DFS

组合型枚举:无限制数量

#include<iostream>
using namespace std;

int a[10000];
int n,sum; 

void dfs(int t,int idx,int s){
	if(s == n){
		cout << n << "=" << a[0];
		for(int i = 1; i < t; i++){
			cout << "+" << a[i];
		}
		cout << endl;
		sum++;
		return;
	}
	for(int i = idx; i < n; i++){
		if(i <= n-s+1){
			a[t] = i;
			dfs(t+1,i,s+i);
		} 
	}
}

int main(){
	cin >> n;
	dfs(0,1,0);	
	return 0;
} 

组合型枚举:有数量限制

#include<bits/stdc++.h>
using namespace std;
int a[1000];
int n,k;

void func(int m,int idx){
	if(m == k){
		for(int i = 0; i < m; i++) printf("%3d",a[i]);
		cout << endl;
	}
	for(int i = idx+1; i <= n; i++){ //idx为我上次找到元素的位置从后面继续找
		a[m] = i;
		func(m+1,i); //m+1搭配方案里的元素+1,idx=i为我现在看到的元素位置
	}
}

int main(){
	cin >> n >> k;
	func(0,0);
	return 0;
}

全排列

#include<iostream>
using namespace std;

int n;
int a[10000],v[1000000];

void dfs(int x){
	if(x == n){
		for(int i = 0; i < n; i++) cout << a[i] << " ";
		cout << endl;
	}
	for(int i = 1; i <= n; i++){
		if(v[i] == 0){
			a[x] = i;
			v[i] = 1;
			dfs(x+1);
			v[i] = 0;
		}
	}
}

int main(){
	cin >> n;
	dfs(0);
}

八皇后

#include<iostream>
using namespace std;

const int N = 1001;
char g[N][N];
bool col[N],dg[N],udg[N]; 
int n,cnt; 

void dfs(int u){
	if(u == n){
		cnt++;
		cout << "No. " << cnt << endl;
		for(int j = 0; j < n; j++){
			for(int i = 0; i < n; i++){
				if(g[j][i] == 'Q') cout << 1 << " ";
				else cout << 0 << " ";
			}
			cout << endl;
		} 
		return ;
	}
	for(int i = 0; i < n; i++){
		if(!col[i] && !dg[u+i] && !udg[n-u+i]){
			g[u][i] = 'Q';
			col[i] = dg[u+i] = udg[n-u+i] = true;
			dfs(u+1);
			col[i] = dg[u+i] = udg[n-u+i] = false;
			g[u][i] = '.';
		}
	}
} 

int main(){
	n = 8;
	for(int i = 0; i < n; i++){
		for(int j = 0; j < n; j++){
			g[i][j] = '.';
		}
	}
	dfs(0);
	return 0;
} 

图论:

        存储与遍历

               vectoer

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;

const int N = 1e3+10;
vector<vector<int> > g(N);
bool st[N];
int n,m; 

void bfs(int x){
	queue<int> q;
	q.push(x);
	st[x] = 1;
	cout << x << " ";
	while(!q.empty()){
		int t = q.front();
		q.pop();
		for(int i = 0; i < g[t].size(); i++){
			if(!st[g[t][i]]){
				q.push(g[t][i]);
				st[g[t][i]] = 1;
				cout << g[t][i] << " ";
			}
		}
	}
}

void dfs(int x){
	st[x] = 1;
	cout << x << " ";
	for(int i = 0; i < g[x].size(); i++){
		if(!st[g[x][i]]) dfs(g[x][i]);
	}
}

int main(){
	cin >> n >> m;
	for(int i = 1; i <= m; i++){
		int a,b; cin >> a >> b;
		g[a].push_back(b);
		g[b].push_back(a);
	}
	for(int i = 1; i <= n; i++){
		for(int j = 0; j < g[i].size(); j++){
			cout << i << "---->" << g[i][j] << endl;
		}
	} 
	bfs(1);
	cout << endl;
	memset(st,0,sizeof(st));
	dfs(1);
	return 0;
} 


/*
9 8
1 2
1 7
1 4
2 8
2 5
4 3
3 9
4 6
*/ 

临接矩阵-数组

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;

const int N = 1e3+10;
int n,m;
int g[N][N];
int v[N];

void bfs(int x){
	queue<int> q;
	q.push(x);
	v[x] = 1;
	cout << x << " ";
	while(!q.empty()){
		int t = q.front();
		q.pop();
		for(int i = 1; i <= n; i++){
			if(g[t][i] && !v[i]){
				q.push(i);
				cout << i << " ";
				v[i] = 1;
			}
		}
	}
}

void dfs(int x){
	v[x] = 1;
	cout << x << " ";
	for(int i = 1; i <= n; i++){
		if(g[x][i] && !v[i]) dfs(i);
	}
}

int main(){
	int x; 
	cin >> n >> m;
	for(int i = 1; i <= m; i++){
		int a,b; cin >> a >> b;
		g[a][b] = g[b][a] = 1;
	}
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			if(g[i][j]) cout << i << "------>" << j << endl;
		}
	} 
	bfs(1); 
	cout << endl;
	memset(v,0,sizeof v);
	dfs(1);
	return 0;
} 


/*
9 8
1 2
1 7
1 4
2 8
2 5
4 3
3 9
4 6
*/ 

临接表-链表

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;

const int N = 1e5 + 10, M = 1e6 + 10;
int st[N];
int n, m;
// h数组存储所有节点的边链表的头节点
// e数组存储的是节点 
int h[N], e[M], ne[M], w[M], idx;

void add(int a, int b) {
	e[idx] = b;
	ne[idx] = h[a];
	h[a] = idx;
	idx++;
}

void dfs(int u) {
	cout << u << " ";
	st[u] = 1;
	for(int i = h[u]; i != -1; i = ne[i]){
		int j = e[i];
		if(!st[j]) dfs(j);
	}
} 

void bfs(int u) {
	queue<int> q;
	q.push(u);
	st[u] = 1;
	cout << u << ' ';
	while(!q.empty()){
		int t = q.front();
		q.pop();
		for(int i = h[t]; i != -1; i = ne[i]){
			int j = e[i];
			if(!st[j]){
				cout << j << " ";
				st[j] = 1;
				q.push(j);
			} 
		} 
	} 
}


int main() {
	memset(h, -1, sizeof h);
	// 链式前向星
	 
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b, c; cin >> a >> b;
		add(a, b);
		add(b, a);
	}
	
	for (int i = 1; i <= n; i++) { // 枚举每一个顶点 
		cout << i << "的所有边有这些:";  // 枚举当前顶点的所有边 
		for (int j = h[i]; j != -1; j = ne[j]) {
			cout << e[j] << " ";
		}
		cout << endl;
	}
	
	dfs(1);
	cout << endl;
	memset(st, 0, sizeof st);
	bfs(1);
	
    return 0;
}



/*
一个示例图: 无向图 
9 8
1 2
1 7
1 4
2 8
2 5
4 3
3 9
4 6
*/


网站公告

今日签到

点亮在社区的每一天
去签到