Secret Poems - HihoCoder - 1632
图一 图二
Following the order indicated by arrows, you can get “THISISAVERYGOODPOEMITHINK”, and that can mean something.But after some time, poets found out that some Yongzheng’s secret agent called “Mr. blood dripping” could read this kind of poems too. That was dangerous. So they introduced a new order of writing poems as shown in figure 2. And they wanted to convert the old poems written in old order as figure1 into the ones in new order. Please help them.InputThere are no more than 10 test cases.For each test case:The first line is an integer N( 1 <= N <= 100), indicating that a poem is a N×N matrix which consist of capital letters.Then N lines follow, each line is an N letters string. These N lines represent a poem in old order.OutputFor each test case, convert the poem in old order into a poem in new order.
Sample Input
5 THSAD IIVOP SEOOH RGETI YMINK 2 AB CD 4 ABCD EFGH IJKL MNOP
Sample Output
THISI POEMS DNKIA OIHTV OGYRE AB DC ABEI KHLF NPOC MJGD
大致题意：
就是先按照图一的方式进行遍历这个图——按照图一的顺序存成一个字符串；然后按照图二的“回字形 ”顺序进行调整，然后输出图二！
AC题解：
没错，题意就是这么简单！写的时候一不留神就出了很多BUG！然后改来改去花费了不少的时间！还是得细心点点!
没有想到简单的办法，只好进行If...else...的嵌套！
1 #include<stdio.h> 2 #include<math.h> 3 #include<string.h> 4 #include<iostream> 5 #include<algorithm> 6 #include<string> 7 #include<set> 8 #include<vector> 9 #define inf 0x3f3f3f3f //Pangu and Stones ,I 10 #define ll long long 11 using namespace std; 12 #define N 108 13 char a[N][N]; 14 char ans[N][N]; 15 bool vis[N][N]; 16 int dir[6][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,-1},{-1,1} }; 17 void factans(string s,int n){ 18 int x=0,y=0; 19 memset(vis,false,sizeof(vis)); 20 vis[0][0]=true; 21 memset(ans,‘6‘,sizeof(ans)); 22 ans[0][0]=s[0]; 23 int op=-1; 24 //先按之前的进行，右下左上右———— 25 for(int i=1;i<s.length();i++){ 26 if(op!=-1){ 27 int dx=x+dir[op][0]; 28 int dy=y+dir[op][1]; 29 if(dx>=0&&dx<=n-1&&dy>=0&&dy<=n-1&&vis[dx][dy]==false){ 30 x=dx;y=dy; 31 ans[x][y]=s[i]; 32 vis[x][y]=true; 33 continue; 34 } 35 } 36 if(y+1<=n-1&&vis[x][y+1]==false){//右下左上右———— 37 ans[x][y+1]=s[i]; 38 y++; 39 op=0;vis[x][y]=true; 40 }else if(x+1<=n-1&&vis[x+1][y]==false){ 41 ans[x+1][y]=s[i]; 42 x++; 43 op=1;vis[x][y]=true; 44 }else if(y-1>=0&&vis[x][y-1]==false){ 45 ans[x][y-1]=s[i]; 46 y--; 47 op=2;vis[x][y]=true; 48 }else if(x-1>=0&&vis[x-1][y]==false){ 49 ans[x-1][y]=s[i]; 50 x--; 51 op=3;vis[x][y]=true; 52 } 53 } 54 for(int i=0;i<n;i++) 55 ans[i][n]=‘\0‘; 56 for(int i=0;i<n;i++) 57 printf("%s\n",ans[i]); 58 59 } 60 int main(){ 61 int n; 62 63 while(scanf("%d",&n)!=EOF){ 64 //int num=1; 65 for(int i=0;i<n;i++) 66 scanf("%s",a[i]); 67 string s; 68 s.clear(); 69 int i=0,j=0; 70 int op=-1; 71 memset(vis,false,sizeof(vis)); 72 while(1){ 73 74 s+=a[i][j];//存储当前位置 75 76 if(i==n-1&&j==n-1){ 77 // cout<<"end: "<<s<<endl; 78 break; 79 } 80 81 vis[i][j]=true; 82 83 if(i==0){//在第一行，可右0，可左下4，可下1 84 if(j+1<=n-1&&op!=0){ 85 i+=dir[0][0]; 86 j+=dir[0][1]; 87 op=0; 88 } 89 else if(i+1<=n-1&&j-1>=0&&vis[i+1][j-1]==false){ 90 i+=dir[4][0]; 91 j+=dir[4][1]; 92 op=4; 93 } 94 else{ 95 i+=dir[1][0]; 96 j+=dir[1][1]; 97 op=1; 98 } 99 } 100 else if(j==0){//在第一列，先下1，后右上5，后右0 101 if(op!=1&&i+1<=n-1){ 102 i+=dir[1][0]; 103 j+=dir[1][1]; 104 op=1; 105 }else if(i-1>=0&&j+1<=n-1&&vis[i-1][j+1]==false){ 106 i+=dir[5][0]; 107 j+=dir[5][1]; 108 op=5; 109 } 110 else{ 111 i+=dir[0][0]; 112 j+=dir[0][1]; 113 op=0; 114 } 115 } 116 else if(i==n-1){//最后一行，先右上5，后右0 117 if(i-1>=0&&j+1<=n-1&&vis[i-1][j+1]==false){ 118 i+=dir[5][0]; 119 j+=dir[5][1]; 120 op=5; 121 } 122 else{ 123 i+=dir[0][0]; 124 j+=dir[0][1]; 125 op=0; 126 } 127 } 128 else if(j==n-1){//最后一列，先左下4，后下1 129 if(i+1<=n-1&&j-1>=0&&vis[i+1][j-1]==false){ 130 i+=dir[4][0]; 131 j+=dir[4][1]; 132 op=4; 133 } 134 else if(op!=1){ 135 i+=dir[1][0]; 136 j+=dir[1][1]; 137 op=1; 138 } 139 } 140 else{//其余的照旧就可以了 141 i+=dir[op][0]; 142 j+=dir[op][1]; 143 } 144 } 145 factans(s,n); 146 147 } 148 return 0; 149 }
