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There is a mobile piece and a stationary piece on the N×M chessboard. The available moves of the mobile piece are the same as set out in the image below. You need to capture the stationary piece by moving the mobile piece with the minimum amount of moves. For each test case, you should output the
minimum number of movements to catch a defending piece at the first line of each
test case. If not moveable, output equals ‘-1’. Input 2 Case #2 5 |
代码：
#include <iostream> #include <stdio.h> #include <queue> #include <string.h> using namespace std; typedef struct { int x; int y; int level; }data; int mv[8][2] = {{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1}}; int main() { //freopen("test.txt","r",stdin); int T; cin>>T; for(int t=1; t<=T; t++) { int n,m,r1,c1,r2,c2; cin>>n>>m; int a[n+1][m+1]; memset(a,0,sizeof(int)*(n+1)*(m+1)); cin>>r1>>c1>>r2>>c2; data d,d1,d2; queue<data> qt; d.x = r1; d.y = c1; d.level = 0; qt.push(d); a[d.x][d.y] = 2; int tmx,tmy,tml; int steps = 0; bool f = false; while(!qt.empty()) { if(f) { break; } d1 = qt.front(); qt.pop(); for(int k=0; k<8; k++) { tmx = d1.x + mv[k][0]; tmy = d1.y + mv[k][1]; tml = d1.level + 1; if(tmx>=1 && tmx<=n && tmy>=1 && tmy<=m && a[tmx][tmy] == 0) { if(tmx == r2 && tmy == c2) { steps = tml; f = true; break; } d2.x = tmx; d2.y = tmy; d2.level = tml; qt.push(d2); a[d2.x][d2.y] = 2; } } } if(!f) { steps = -1; } cout<<"Case #"<<t<<endl; cout<<steps<<endl; } //cout << "Hello world!" << endl; return 0; }
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原文地址：http://www.cnblogs.com/kingshow123/p/practicec2.html
