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HDU 4121 Xiangqi --模拟
题意： 给一个象棋局势，问黑棋是否死棋了，黑棋只有一个将，红棋可能有2~7个棋，分别可能是车，马，炮以及帅。
解法： 开始写法是对每个棋子，都处理处他能吃的地方，赋为-1，然后判断将能不能走到非-1的点。但是WA了好久，也找不出反例，但就是觉得不行，因为可能有将吃子的情况，可能有hack点。但是比赛后还是被我调出来了。
代码：
#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>using namespace std;#define N 1017int chess[14][13],mp[14][13]; //0 : Non 1:G帅 2:R车 3:H Horse 4:C Cannonint dx[4] = {0,-1,0,1};int dy[4] = {1,0,-1,0};bool InPalace(int nx,int ny) { if(nx >= 1 && nx <= 3 && ny >= 4 && ny <= 6) return true; return false; }bool InChess(int nx,int ny) { if(nx >= 1 && nx <= 10 && ny >= 1 && ny <= 9) return true; return false;}int main(){ int n,X,Y,i,j,k; int x,y; char ss[4]; while(scanf("%d%d%d",&n,&X,&Y)!=EOF && (n+X+Y)) { memset(chess,0,sizeof(chess)); memset(mp,0,sizeof(mp)); for(i=1;i<=n;i++) { //0 : Non 1:G帅 2:R车 3:H Horse 4:C Cannon scanf("%s%d%d",ss,&x,&y); if(ss[0] == ‘G‘) chess[x][y] = 1; else if(ss[0] == ‘R‘) chess[x][y] = 2; else if(ss[0] == ‘H‘) chess[x][y] = 3; else if(ss[0] == ‘C‘) chess[x][y] = 4; } for(i=1;i<=10;i++) { for(j=1;j<=9;j++) { if(chess[i][j] == 1) // shuai { for(k=i-1;k>=1;k--) { if(chess[k][j] != 0) { mp[k][j] = -1; break; } if(k <= 3 && chess[k][j] == 0) mp[k][j] = -1; } } else if(chess[i][j] == 2) // R che { for(int D=0;D<4;D++) { int kx = i, ky = j; while(1) { kx = kx + dx[D]; ky = ky + dy[D]; if(!InChess(kx,ky)) break; if(chess[kx][ky] == 0) mp[kx][ky] = -1; else { mp[kx][ky] = -1; break; } } } } else if(chess[i][j] == 3) //Horse { if(InChess(i-1,j) && chess[i-1][j] == 0 && (i-1 != X || j != Y)) //UP not blocked { if(InChess(i-2,j-1)) mp[i-2][j-1] = -1; if(InChess(i-2,j+1)) mp[i-2][j+1] = -1; } if(InChess(i+1,j) && chess[i+1][j] == 0 && (i+1 != X || j != Y)) //DOWN not blocked { if(InChess(i+2,j-1)) mp[i+2][j-1] = -1; if(InChess(i+2,j+1)) mp[i+2][j+1] = -1; } if(InChess(i,j+1) && chess[i][j+1] == 0 && (i != X || j+1 != Y)) //RIGHT not blocked { if(InChess(i-1,j+2)) mp[i-1][j+2] = -1; if(InChess(i+1,j+2)) mp[i+1][j+2] = -1; } if(InChess(i,j-1) && chess[i][j-1] == 0 && (i != X || j-1 != Y)) //LEFT not blocked { if(InChess(i-1,j-2)) mp[i-1][j-2] = -1; if(InChess(i+1,j-2)) mp[i+1][j-2] = -1; } } else if(chess[i][j] == 4) //Cannon pao { for(int D=0;D<4;D++) { int kx = i, ky = j; int cnt = 0; while(1) { kx = kx + dx[D]; ky = ky + dy[D]; if(!InChess(kx,ky)) break; if(cnt == 1 && chess[kx][ky] == 0) mp[kx][ky] = -1; if(chess[kx][ky] != 0) { if(cnt == 0) cnt++; else if(cnt == 1) { mp[kx][ky] = -1; break; } } } } } } } int tag = 0; for(k=0;k<4;k++) { int kx = X + dx[k]; int ky = Y + dy[k]; if(!InChess(kx,ky) || !InPalace(kx,ky)) continue; if(mp[kx][ky] != -1) { tag = 1; break; } //可以走 } if(tag) puts("NO"); else puts("YES"); } return 0;}
比赛中后来换了一种写法，枚举将能走的四个位置，然后判断此时是否能被吃掉。 如果没有一个安全的地方，那么就死棋了。 这种写起来就好些多了。
代码：
#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>using namespace std;#define N 1017int chess[14][13],mp[14][13]; //0 : Non 1:G帅 2:R车 3:H Horse 4:C Cannonint dx[4] = {0,-1,0,1};int dy[4] = {1,0,-1,0};bool InPalace(int nx,int ny){ if(nx >= 1 && nx <= 3 && ny >= 4 && ny <= 6) return true; return false;}bool InChess(int nx,int ny){ if(nx >= 1 && nx <= 10 && ny >= 1 && ny <= 9) return true; return false;}struct node{ int x,y,type;}q[10];bool Check(int X,int Y,int i,int j,int type) //将死则 return false !{ int k; if(type == 1) //帅 { if(j != Y) return true; //不是一行 for(k=i-1;k>X;k--) { if(chess[k][j] != 0) break; } if(k == X) return false; //老倌子见面,gg return true; } else if(type == 2) //R { int tag = 1; for(int D=0;D<4;D++) { int kx = i, ky = j; while(1) { kx = kx + dx[D]; ky = ky + dy[D]; if(!InChess(kx,ky)) break; if(kx == X && ky == Y) //车能碰到将 { tag = 0; break; } if(chess[kx][ky] != 0) break; } } if(!tag) return false; return true; } else if(type == 3) //Horse { int tag = 1; if(InChess(i-1,j) && chess[i-1][j] == 0 && (i-1 != X && j != Y)) //UP not blocked { if(InChess(i-2,j-1) && i-2 == X && j-1 == Y) tag = 0; if(InChess(i-2,j+1) && i-2 == X && j+1 == Y) tag = 0; } if(InChess(i+1,j) && chess[i+1][j] == 0 && (i+1 != X && j != Y)) //DOWN not blocked { if(InChess(i+2,j-1) && i+2 == X && j-1 == Y) tag = 0; if(InChess(i+2,j+1) && i+2 == X && j+1 == Y) tag = 0; } if(InChess(i,j+1) && chess[i][j+1] == 0 && (i != X && j+1 != Y)) //RIGHT not blocked { if(InChess(i-1,j+2) && i-1 == X && j+2 == Y) tag = 0; if(InChess(i+1,j+2) && i+1 == X && j+2 == Y) tag = 0; } if(InChess(i,j-1) && chess[i][j-1] == 0 && (i != X && j-1 != Y)) //LEFT not blocked { if(InChess(i-1,j-2) && i-1 == X && j-2 == Y) tag = 0; if(InChess(i+1,j-2) && i+1 == X && j-2 == Y) tag = 0; } if(!tag) return false; return true; } else if(type == 4) //Cannon { int tag = 1; for(int D=0;D<4;D++) { int kx = i, ky = j; int cnt = 0; while(1) { kx = kx + dx[D]; ky = ky + dy[D]; if(!InChess(kx,ky)) break; if(cnt == 1 && kx == X && ky == Y) { tag = 0; break; } if(chess[kx][ky] != 0) cnt++; } } if(!tag) return false; return true; } return false;}int main(){ int n,X,Y,i,j,k; int x,y; char ss[4]; while(scanf("%d%d%d",&n,&X,&Y)!=EOF && (n+X+Y)) { memset(chess,0,sizeof(chess)); int tot = 0; for(i=1;i<=n;i++) { //0 : Non 1:G帅 2:R车 3:H Horse 4:C Cannon scanf("%s %d%d",ss,&x,&y); if(ss[0] == ‘G‘) chess[x][y] = 1; else if(ss[0] == ‘R‘) chess[x][y] = 2; else if(ss[0] == ‘H‘) chess[x][y] = 3; else if(ss[0] == ‘C‘) chess[x][y] = 4; node now; now.x = x, now.y = y, now.type = chess[x][y]; q[++tot] = now; } int flag[10]; int tag = 0; for(k=0;k<4;k++) { int nx = X + dx[k]; int ny = Y + dy[k]; memset(flag,0,sizeof(flag)); if(!InChess(nx,ny) || !InPalace(nx,ny)) continue; for(i=1;i<=tot;i++) { if(nx == q[i].x && ny == q[i].y) { flag[i] = 1; chess[nx][ny] = 0; } } int smalltag = 1; for(i=1;i<=tot;i++) { if(flag[i]) continue; bool res = Check(nx,ny,q[i].x,q[i].y,q[i].type); if(!res) { smalltag = 0; break; } } if(smalltag) { tag = 1; break; } // 还原 for(i=1;i<=tot;i++) { if(flag[i]) chess[q[i].x][q[i].y] = q[i].type; } } if(tag) puts("NO"); else puts("YES"); } return 0;}
HDU 4121 Xiangqi --模拟
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