HDU 1372 (BFS)

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BFS：

起点入队；

开始搜索：

读取队首并出队，搜索范围有四周八个点（有的不存在），搜索到的点入队，并将歩长+1；直到读取的点等于终点；

15MS372K

code：

#include <iostream>#include <string>#include <cstdio>#include <queue>using namespace std;struct point{int x;int y;int step;}st,ed;int x1,y1,x2,y2;bool visit[10][10];int xx[8] = {1,2,1,2,-1,-2,-1,-2};int yy[8] = {2,1,-2,-1,2,1,-2,-1};void bfs(){memset(visit,false,sizeof(visit));st.x = x1;st.y = y1;st.step = 0;queue <point> p;p.push(st);while(!p.empty()){ed = p.front();p.pop();if(ed.x == x2 && ed.y == y2)break;for(int i = 0; i < 8; i++){st.x = ed.x + xx[i];st.y = ed.y + yy[i];st.step = ed.step + 1;if(!visit[st.x][st.y]&&st.x>0&&st.x<9&&st.y>0&&st.y<9){visit[st.x][st.y] = true;p.push(st);}}}}int main(){char c1,c2,c;while(scanf("%c%d%c%c%d",&c1,&y1,&c,&c2,&y2)!=EOF){x1 = c1 - 'a' + 1;x2 = c2 - 'a' + 1;bfs();printf("To get from %c%d to %c%d takes %d knight moves.\n",c1,y1,c2,y2,ed.step);getchar();}}

DFS：

由起点对每个点进行周围8个方位访问，递归调用DFS；复杂度较高；本题只有8*8，所以能接受；

375MS332K

code：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <iomanip>using namespace std;int r[8][2]={{2,1},{1,2},{-1,2},{-2,1}, {2,-1},{1,-2},{-1,-2},{-2,-1} };int num[10][10];void dfs(int x1, int y1, int move){if(x1<=0 || x1>=9 || y1<=0 || y1>=9||move>=num[x1][y1])return ;num[x1][y1] = move;for(int i = 0; i < 8; i++){dfs(x1+r[i][0], y1+r[i][1],move+1);}}int main(int argc, char *argv[]){int x1,y1,x2,y2;char c1,c2,c;while(scanf("%c%d%c%c%d",&c1,&y1,&c,&c2,&y2)!=EOF){x1 = c1 - 'a' + 1;x2 = c2 - 'a' + 1;memset(num,100,sizeof(num));dfs(x1,y1,0);printf("To get from %c%d to %c%d takes %d knight moves.\n",c1,y1,c2,y2,num[x2][y2]);getchar();}return 0;}