[noip2011 d1t3] Mayan游戏

无脑码农题==

烦的是很多细节，把几个主要过程分开写，化整为零会清楚很多

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define mxcol 12
int map[10][10];
int goal;
struct data{
    int x,y,w;
}ans[10];
int read(){
    int x=0;
    char ch=getchar();
    while (ch<'0'||ch>'9') ch=getchar();
    while (ch>='0'&&ch<='9'){
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x;
}
bool empty(){
    for (int i=0;i<5;i++)
      for (int j=0;j<7;j++)
        if (map[i][j]) return 0;
    return 1;
}
void DROP(){
    int time[10][10];
    memset(time,-1,sizeof(time));
    for (int x=0;x<5;x++){
        int num=0;
        for (int y=0;y<7;y++)
          if (map[x][y])
            time[x][num++]=y;
    }
    for(int x=0;x<5;x++)
      for (int y=0;y<7;y++)
        map[x][y]=time[x][y]==-1?0:map[x][time[x][y]];
}
bool CLEAR(){
    bool flag=0;
    //横
    for (int i=0;i<3;i++)
      for (int j=0;j<7;j++)
        if (map[i][j]){
            int x;
            for (x=i;x+1<5&&map[i][j]==map[x+1][j];x++);
            if (x-i>=2){
                int nx;
                for (nx=i;nx<=x;nx++){
                    int up=j,down=j;
                    while (up+1<7&&map[nx][up+1]==map[i][j]) up++;
                    while (down-1>=0&&map[nx][down-1]==map[i][j]) down--;
                    if (up-down>=2){
                        for (int ny=down;ny<=up;ny++)
                          map[nx][ny]=0;
                    }
                }
                for (nx=i;nx<=x;nx++)
                  map[nx][j]=0;
                flag=1;
            }
        }
    //竖 
    for (int i=0;i<5;i++)
      for (int j=0;j<5;j++)
        if (map[i][j]){
            int y;
            for (y=j;y+1<7&&map[i][y+1]==map[i][j];y++);
            if (y-j>=2){
                int ny;
                for (ny=j;ny<=y;ny++){
                    int left=i,right=i;
                    while (left>0&&map[left-1][ny]==map[i][j]) left--;
                    while (right+1<7&&map[right+1][ny]==map[i][j]) right++;
                    if (right-left>=2){
                        for (int nx=left;nx<=right;nx++)
                        map[nx][ny]=0;
                    }
                }
                for (ny=j;ny<=y;ny++)
                  map[i][ny]=0;
                flag=1;
            }
        }
    return flag;
}
void dfs(int step){
    if (step>goal){
        if (empty()){//判断游戏结束 
            for (int i=1;i<=goal;i++)
              if (ans[i].w)
                printf("%d %d -1
",ans[i].x+1,ans[i].y);
              else 
                printf("%d %d 1
",ans[i].x,ans[i].y); 
            exit(0);
        }
        return;
    }
    
    int time[mxcol];
    memset(time,0,sizeof(time));
    for (int i=0;i<5;i++)
      for (int j=0;j<7;j++)
        time[map[i][j]]++;
    for (int i=1;i<=10;i++)
      if (time[i]!=0&&time[i]<3) return;//剪枝:当某种颜色块数小于3时必然无解 
    
    for (int x=0;x<4;x++)//优先向右换;按字典序枚举 
      for (int y=0;y<7;y++)
        if (map[x][y]!=map[x+1][y]){//两方块颜色不同时才交换 
             ans[step].x=x,ans[step].y=y;
             ans[step].w=!map[x][y];//枚举的时候是从左往右，但也可能左边空右边有方块，此时标记将右边的方块向左移动 
             int tmp[10][10];
             memcpy(tmp,map,sizeof(tmp));
             swap(map[x][y],map[x+1][y]);
             
             DROP();
             while (CLEAR()) DROP();
             //两个主要过程：上面的方块掉落；合法的方块消去 
             dfs(step+1);
             
             ans[step].x=0,ans[step].y=0;
             ans[step].w=0;
             memcpy(map,tmp,sizeof(map));
        }
}
int main(){
    goal=read();
    memset(map,0,sizeof(map));
    for (int i=0;i<5;i++){
        for (int j=0;;j++){
            int u=read();
            if (!u) break;
            map[i][j]=u;
        }
    }
    dfs(1);
    printf("-1
");
    return 0;
}