Quick test-Basic Search (Hangzhou Telecom 5)

HDU 1241 oil deposits

Water question, BFs, determine the number of blocks in the region.

Code List:

#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 questions, BFs (DFS), sum of accessible points.

#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 bonedfs, which has learned parity pruning.

Q: there is only one door in the Maze and the door is opened only in the K-second. Can you walk out of the maze.

# 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); // parity pruning: judge <= 0 and an even number can continue 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", & 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"); // The number of walk points must be greater than 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 priority queue + BFs. Note that there are multiple rescue sites, so you only need to go to the nearest rescue site from the origin point.

#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

Priority queue + BFS + path restoration.

# 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]; // path of the B array record: int n, m; int vis [105] [105]; char s [105] [105]; int XY [4] [2] = {0,-1 }}; 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 ", & N, & M )! = EOF) {for (INT I = 0; I <n; I ++) scanf ("% s", s [I]); memset (VIS, 0, sizeof (VIS); BFS ();} return 0 ;}

Quick test-Basic Search (Hangzhou Telecom 5)