POJ 2243 bfs

题意：求棋子从起点到终点的最小步数。

分析：bfs。

注意：用while(scanf("%s%s",str1,str2)!=EOF)会tle。这边2次tle后来改成while(cin>>str1>>str2)就过了。

收获：bfs实在是弱，这题几乎是按照自己思路，一点一点敲出来的，不错。继续加油。（这题以前做过，当时是用数组模拟队列，是照别人做的。）

View Code

// I'm the Topcoder//C#include <stdio.h>#include <stdlib.h>#include <string.h>#include <ctype.h>#include <math.h>#include <time.h>//C++#include <iostream>#include <algorithm>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <cctype>#include <stack>#include <string>#include <list>#include <queue>#include <map>#include <vector>#include <deque>#include <set>using namespace std;//*************************OUTPUT*************************#ifdef WIN32#define INT64 "%I64d"#define UINT64 "%I64u"#else#define INT64 "%lld"#define UINT64 "%llu"#endif//**************************CONSTANT***********************#define INF 0x3f3f3f3f#define eps 1e-8#define PI acos(-1.)#define PI2 asin (1.);typedef long long LL;//typedef __int64 LL;   //codeforcestypedef unsigned int ui;typedef unsigned long long ui64;#define MP make_pairtypedef vector<int> VI;typedef pair<int, int> PII;#define pb push_back#define mp make_pair//***************************SENTENCE************************#define CL(a,b) memset (a, b, sizeof (a))#define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b))#define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c))//****************************FUNCTION************************template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); }template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; }template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; }// aply for the memory of the stack//#pragma comment (linker, "/STACK:1024000000,1024000000")//endconst int maxn = 110;char str1[3];char str2[3];//int map[10][10];struct node{    int x;    int y;    //int time;};queue<node> Q;int sx,sy;int dir[8][2]={{-1,-2},{-2,-1},{-2,1},{-1,2},{1,-2},{2,-1},{2,1},{1,2}};//int dir[8][2]={{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1},{-2,1}};int dis[maxn][maxn];int vis[maxn][maxn];void bfs(node s){    Q.push(s);   // dis[s.x][s.y]=0;    //vis[s.x][s.y]=1;   // node hd;    while(!Q.empty()){        //printf("&&&&\n");        node temp=Q.front();        Q.pop();        int x=temp.x;        int y=temp.y;        vis[x][y]=1;//        if(x==sx&&y==sy)//边界//        {//            printf("%d\n",dis[sx][sy]);//            //printf("sx=%d  sy=%d\n",sx,sy);//            return;//        }        for(int i=0;i<8;i++){            int xx=x+dir[i][0];            int yy=y+dir[i][1];            if(xx>=0&&xx<8&&yy>=0&&yy<8&&!vis[xx][yy]){                dis[xx][yy]=dis[x][y]+1;                temp.x=xx;                temp.y=yy;                Q.push(temp);                vis[xx][yy]=1;            }        }    }    //if()    printf("To get from %s to %s takes %d knight moves.\n",str1,str2,dis[sx][sy]);       // printf("%d\n",dis[sx][sy]);}int main(){    while(cin>>str1>>str2){        //node edge;        memset(vis,0,sizeof(vis));        memset(dis,0,sizeof(dis));        int startx=str1[0]-'a';//开始起点        int starty=str1[1]-'1';       // printf("startx=%d starty=%d\n",startx,starty);        sx=str2[0]-'a';//终点        sy=str2[1]-'1';        //printf("sx=%d sy=%d\n",sx,sy);        dis[startx][starty]=0;        node s;        vis[startx][starty]=1;        s.x=startx;        s.y=starty;        //s.time=0;        if(startx==sx&&starty==sy){            printf("To get from %s to %s takes 0 knight moves.\n",str1,str2);            continue;        }       else  bfs(s);    }    return 0;}