小试牛刀-搜索基础入门（杭电五题）
hdu 1241 Oil Deposits
水题，BFS，判断区域的块数。
代码清单：
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef pair<int,int>P;
int m,n;
char s[105][105];
int xy[8][2]={{0,-1},{-1,-1},{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1}};
void bfs(int x,int y){
queue<P>q;
s[x][y]='*';
q.push(P(x,y));
while(q.size()){
P p=q.front(); q.pop();
for(int i=0;i<8;i++){
int xx=p.first+xy[i][0];
int yy=p.second+xy[i][1];
if(xx>=0&&xx<m&&yy>=0&&yy<n&&s[xx][yy]!='*'){
s[xx][yy]='*';
q.push(P(xx,yy));
}
}
}
}
int main(){
while(scanf("%d%d",&m,&n)!=EOF){
if(m==0&&n==0) break;
//memset(s,'\0',sizeof(s));
for(int i=0;i<m;i++)
scanf("%s",s[i]);
int ans=0;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(s[i][j]=='@'){
ans++;
bfs(i,j);
}
}
}
printf("%d\n",ans);
}return 0;
}
hdu1312 Red and Black
水题，BFS（DFS），求可走点的总和。#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef pair<int,int>P;
int m,n;
int sx,sy;
int vis[25][25];
char s[25][25];
int xy[4][2]={{0,-1},{-1,0},{0,1},{1,0}};
int bfs(){
int sum=0;
queue<P>q;
while(q.size()) q.pop();
q.push(P(sx,sy));
vis[sx][sy]=1;
while(q.size()){
P p=q.front();
q.pop();
sum++;
for(int i=0;i<4;i++){
int xx=p.first+xy[i][0];
int yy=p.second+xy[i][1];
if(xx>=0&&xx<n&&yy>=0&&yy<m&&s[xx][yy]!='#'&&!vis[xx][yy]){
vis[xx][yy]=1; //cout<<xx<<" "<<yy<<endl;
q.push(P(xx,yy));
}
}
}return sum;
}
int main(){
while(scanf("%d%d",&m,&n)!=EOF){
if(m==0&&n==0) break;
for(int i=0;i<n;i++){
scanf("%s",s[i]);
for(int j=0;j<m;j++){
if(s[i][j]=='@'){
sx=i;
sy=j;
}
}
}
memset(vis,0,sizeof(vis));
printf("%d\n",bfs());
}return 0;
}
hdu 1010 Tempter of the Bone
DFS，从中学到了奇偶性剪枝。题意：迷宫中只有一扇门，且只在第K秒时开门，问能否走出迷宫。
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int first;
int N,M,T;
int sx,sy;
int ex,ey;
char s[8][8];
int x[4]={1,-1,0,0};
int y[4]={0,0,-1,1};
void dfs(int X,int Y,int t)
{
if(t==T){
if(X==ex&&Y==ey)
first=1;
return ;
}
if(first) return ;
int judge=abs(X-ex)+abs(Y-ey)-abs(t-T); //奇偶性剪枝:judge<=0且为偶数才可以继续
if(judge>0 || judge%2) return ;
for(int i=0;i<4;i++){
int xx=X+x[i];
int yy=Y+y[i];
if(xx>=0&&xx<N&&yy>=0&&yy<M&&s[xx][yy]!='X'){
s[xx][yy]='X';
dfs(xx,yy,t+1);
s[xx][yy]='.';
}
}
}
int main()
{
while(scanf("%d%d%d",&N,&M,&T)!=EOF){
if(N==0&&M==0&&T==0) break;
int pos=0;
for(int i=0;i<N;i++){
scanf("%s",s[i]);
for(int j=0;j<M;j++){
if(s[i][j]=='S'){
sx=i;
sy=j;
}
else if(s[i][j]=='D'){
ex=i;
ey=j;
}
else if(s[i][j]=='X'){
pos++;
}
}
}
if(N*M-pos<=T) printf("NO\n"); //可走的点数必须大于T
else{
first=0;
s[sx][sy]='X';
dfs(sx,sy,0);
if(first) printf("YES\n");
else printf("NO\n");
}
}return 0;
}
hdu 1242 Rescue
优先队列+BFS。注意救援有多个，所以以终点为起始点，只要走到一个最近的救援点即可。#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct edge{
int x,y;
int t;
friend bool operator<(edge a,edge b){
return a.t>b.t;
}
};
int N,M;
int ex,ey,ok;
int vis[205][205];
char s[205][205];
int xy[4][2]={{0,-1},{-1,0},{0,1},{1,0}};
int bfs(){
priority_queue<edge>q;
while(q.size()) q.pop();
edge p;
p.x=ex; p.y=ey; p.t=0;
vis[ex][ey]=1;
q.push(p);
while(q.size()){
p=q.top();
q.pop();
if(s[p.x][p.y]=='r'){
return p.t;
}
for(int i=0;i<4;i++){
edge w;
w.x=p.x+xy[i][0];
w.y=p.y+xy[i][1];
if(w.x>=0&&w.x<N&&w.y>=0&&w.y<M&&s[w.x][w.y]!='#'&&!vis[w.x][w.y]){
w.t=p.t+1;
if(s[w.x][w.y]=='x') w.t++;
vis[w.x][w.y]=1;
q.push(w);
}
}
}
return -1;
}
int main(){
while(scanf("%d%d",&N,&M)!=EOF){
for(int i=0;i<N;i++){
scanf("%s",s[i]);
for(int j=0;j<M;j++){
if(s[i][j]=='a'){
ex=i;
ey=j;
}
}
}
memset(vis,0,sizeof(vis));
int ans=bfs();
if(ans!=-1) printf("%d\n",ans);
else printf("Poor ANGEL has to stay in the prison all his life.\n");
}return 0;
}
hdu 1026 Ignatius and the Princess I
优先队列+BFS+路径还原。
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct edge
{
int x,y;
int t,r;
friend bool operator<(edge c,edge d){
return c.t>d.t;
}
}a[10000],b[105][105]; //b数组记录路径
int N,M;
int vis[105][105];
char s[105][105];
int xy[4][2]={{0,-1},{-1,0},{0,1},{1,0}};
void bfs()
{
priority_queue<edge>q;
while(q.size()) q.pop();
edge p,w;
p.x=0; p.y=0; p.t=0,p.r=0;
vis[0][0]=1;
q.push(p);
int first=0;
while(q.size())
{
p=q.top(); q.pop();
if(p.x==N-1&&p.y==M-1){
first=1;
break;
}
for(int i=0;i<4;i++){
w.x=p.x+xy[i][0];
w.y=p.y+xy[i][1];
if(w.x>=0&&w.x<N&&w.y>=0&&w.y<M&&s[w.x][w.y]!='X'&&!vis[w.x][w.y]){
b[w.x][w.y].x=p.x;
b[w.x][w.y].y=p.y;
w.t=p.t+1;
w.r=p.r+1;
vis[w.x][w.y]=1;
if(s[w.x][w.y]!='.')
w.t=w.t+s[w.x][w.y]-'0';
q.push(w);
}
}
}
if(first)
{
int k=1;
int i=p.r-1,xx=p.x,yy=p.y;
a[p.r].x=p.x; a[p.r].y=p.y;
for(i=p.r-1;i>=0;i--){
a[i].x=b[xx][yy].x;
a[i].y=b[xx][yy].y;
xx=a[i].x;
yy=a[i].y;
}
printf("It takes %d seconds to reach the target position, let me show you the way.\n",p.t);
for(int i=0;i<p.r;i++){
printf("%ds:(%d,%d)->(%d,%d)\n",k++,a[i].x,a[i].y,a[i+1].x,a[i+1].y);
if(s[a[i+1].x][a[i+1].y]!='.'&&s[a[i+1].x][a[i+1].y]!='X'){
int tt=s[a[i+1].x][a[i+1].y]-'0';
while(tt--) printf("%ds:FIGHT AT (%d,%d)\n",k++,a[i+1].x,a[i+1].y);
}
}
}
else printf("God please help our poor hero.\n");
printf("FINISH\n");
}
int main()
{
while(scanf("%d%d",&N,&M)!=EOF){
for(int i=0;i<N;i++)
scanf("%s",s[i]);
memset(vis,0,sizeof(vis));
bfs();
}return 0;
}