BFS(广度优先搜索)简单例题（二）
本篇博客是关于BFS（广度优先搜索）问题的练习。
填涂颜色 洛谷 P1162
题意：由数字00组成的方阵中，有一任意形状闭合圈，闭合圈由数字11构成，围圈时只走上下左右44个方向。现要求把闭合圈内的所有空间都填写成22.
题解：本题如果搜索圈内的1，非常的麻烦但是如果搜索圈外的1，那句比较简单点了，但是在搜索的时候并不是圈外所有的点都是连通的，所以可以暴力搜索圈外的点，不管圈外的点是不是孤立的，一定能和四个边界相连通的所以，从边的范围遍历。
#include
#include
#include
#include
using namespace std;
#define N 100
struct Node{
int x, y;
} p, tmp;
queueq;
int vis[N][N];
int dis[N][N];
int dx[4] ={1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int n;
void Bfs(int st, int en){
memset(vis, 0, sizeof(vis));
p.x = st;
p.y = en;
if(dis[st][en] != 1 && dis[st][en] != 3){
if(dis[st][en] == 0)dis[p.x][p.y] = 3;
vis[p.x][p.y] = 1;
q.push(p);
int cn;
while(!q.empty()){
p = q.front();
q.pop();
for(int i = 0; i < 4; i++){
tmp.x = p.x + dx[i];
tmp.y = p.y + dy[i];
if(tmp.x >= 0 && tmp.x < n && tmp.y >= 0 && tmp.y < n && !vis[tmp.x][tmp.y] && !dis[tmp.x][tmp.y]){
q.push(tmp);
vis[tmp.x][tmp.y] = 1;
dis[tmp.x][tmp.y] = 3;
}
}
}
}
}
int main(){
while(~scanf("%d", &n)){
while(!q.empty())q.pop();
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
cin>>dis[i][j];
}
}
for(int i = 0; i < n; i++){
Bfs(i, 0);
Bfs(0, i);//遍历所用边界点
Bfs(n-1, i);
Bfs(i, n-1);
}
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
if(dis[i][j] == 3)
cout<<0<<" ";
else if(dis[i][j] == 0)
cout<<2<<" ";
else cout< Battle City POJ 2312
链接：http://poj.org/problem?id=2312
#include
#include
#include
#include
#include
using namespace std;
#define N 305
struct Node{
int x, y, step;
friend bool operator<(Node A, Node B){
return A.step > B.step;
}
}tmp1, tmp2;
int vis[N][N];
char a[N][N];
int n, m, k, flag;
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1,-1, 0, 0};
void bfs(int st, int en){
memset(vis, 0, sizeof(vis));
vis[st][en] = 1;
tmp1.x = st;
tmp1.y = en;
tmp1.step = 0;
priority_queueq;
q.push(tmp1);
while(!q.empty()){
tmp1 = q.top();
q.pop();
if(a[tmp1.x][tmp1.y] == 'T'){
flag = 1;
k = tmp1.step;
break;
}
//cout<=0 && tmp2.x< n&&tmp2.y >= 0&&tmp2.y < m && !vis[tmp2.x][tmp2.y] && a[tmp2.x][tmp2.y] != 'S'&&a[tmp2.x][tmp2.y] != 'R'){
if(a[tmp2.x][tmp2.y] == 'B'){//如果遇到可以涉及的砖，回合加2
tmp2.step = tmp1.step+2;
}
else{
tmp2.step = tmp1.step+1;//其他情况回合加1
}
vis[tmp2.x][tmp2.y] = 1;
q.push(tmp2);
}
}
}
}
int main(){
int st, en;
while(scanf("%d %d", &n, &m) && (m != 0 && n != 0)){
for(int i = 0; i < n; i++){
scanf("%s", a[i]);
for(int j = 0; j < m; j++){
if(a[i][j] == 'Y'){
st = i;
en = j;
}
}
}
flag = 0;
bfs(st, en);
if(!flag)cout<<-1< 泉水 HRBUST 1143
链接：http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=1143
#include
#include
#include
#include
#include
using namespace std;
#define N 1005
struct Node {
int x, y, pre;
}tmp1, tmp2;
int vis[N][N];
int a[N][N];
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1,-1, 0, 0};
int n, m, step;
void bfs(int st, int en, int pre){
memset(vis, 0, sizeof(vis));
vis[st][en] = 1;
tmp1.x = st;
tmp1.y = en;
step = 0;
tmp1.pre = pre;
queueq;
q.push(tmp1);
while(!q.empty()){
tmp1 = q.front();
q.pop();
//cout< 0 && tmp2.x <= n && tmp2.y > 0 &&tmp2.y <= m && !vis[tmp2.x][tmp2.y] && a[tmp2.x][tmp2.y] <= tmp1.pre){
vis[tmp2.x][tmp2.y] = 1;
step++;
tmp2.pre = tmp1.pre;
//cout<>a[i][j];
}
}
pre = a[st][en];
//cout< 马的遍历 洛谷 P1443
链接：https://www.luogu.org/problemnew/show/P1443
#include
#include
#include
#include
#include
using namespace std;
#define N 405
struct Node{
int x, y;
}tmp1, tmp2;
queueq;
int dx[8] = {1,-1,2,-2,-1,1,2,-2};
int dy[8] = {2,2,1,1,-2,-2,-1,-1};
int vis[N][N];
int dis[N][N];
int n, m;
void Bfs(int st, int en){
memset(vis, 0, sizeof(vis));
memset(dis, 0, sizeof(vis));
tmp1.x = st;
tmp1.y = en;
vis[tmp1.x][tmp1.y] = 1;
q.push(tmp1);
while(!q.empty()){
tmp1 = q.front();
q.pop();
for(int i = 0; i < 8; i++){
tmp2.x = tmp1.x + dx[i];
tmp2.y = tmp1.y + dy[i];
if(tmp2.x >= 1 && tmp2.x <= n && tmp2.y >= 1 && tmp2.y <= m && !vis[tmp2.x][tmp2.y]){
q.push(tmp2);
dis[tmp2.x][tmp2.y] = dis[tmp1.x][tmp1.y] + 1;
vis[tmp2.x][tmp2.y] = 1;
}
}
}
}
int main(){
int st, en;
cin>>n>>m>>st>>en;
Bfs(st, en);
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
if(dis[i][j] || (i == st && j == en))
printf("%-5d",dis[i][j]);
else printf("%-5d", -1);
}
cout< Rescue HDU 1242
链接：http://acm.hdu.edu.cn/showproblem.php?pid=1242
#include
#include
#include
#include
#include
#define N 205
using namespace std;
struct Node{
int x, y, val;
friend bool operator<(Node A, Node B){
return A.val > B.val;
}
}tmp1, tmp2;
priority_queueq;
int dx[]= {-1,0,0,1};
int dy[]= {0,-1,1,0};
int vis[N][N];
char a[N][N];
int n, m, flag, k;
void bfs(int st, int en){
memset(vis, 0, sizeof(vis));
vis[st][en] = 1;
tmp1.x = st;
tmp1.y = en;
tmp1.val = 0;
q.push(tmp1);
while(!q.empty()){
tmp1 = q.top();
q.pop();
if(a[tmp1.x][tmp1.y] == 'r'){
k = tmp1.val;
flag = 1;
break ;
}
// cout<= n || tmp2.x < 0 || tmp2.y < 0 || tmp2.y >= m || vis[tmp2.x][tmp2.y]|| a[tmp2.x][tmp2.y] == '#')
continue;
if(a[tmp2.x][tmp2.y] == 'x'){
tmp2.val = tmp1.val + 2;
q.push(tmp2);
}
else {
tmp2.val = tmp1.val + 1;
q.push(tmp2);
}
vis[tmp2.x][tmp2.y] = 1;
}
}
}
int main(){
int st, en;
while(~scanf("%d %d", &n, &m)){
while(!q.empty())q.pop();
for(int i = 0; i < n; i++){
scanf("%s", a[i]);
}
flag = 0;
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(a[i][j] =='a'){
st = i;
en = j;
}
}
}
bfs(st, en);
if(!flag)cout<<"Poor ANGEL has to stay in the prison all his life."<