HDU 4328 Cut the cake(DP)
没早发现这个DP，一直纠结各种图论题，蛋疼。。无奈水平太菜，最后两个小时都没搞出，本来想开10个标记数组的。。。。搞到最后半小时，发现同颜色的不会算，暴力吧。。。然后有个小细节没搞好，计算的时候。WA了，无奈还有10分钟，唉，已无力，查数据。。吃完饭，查出数据。。。过了。。。200+的DP，写的纠结。。。这次又是被虐了。。。QC只出了一个线段树模版，这个没做出，不应该啊，感觉数据应该不会很强，不会乱搞啊。。。
1 #include <stdio.h> 2 #include <string.h> 3 int rr[1001][1001],rc[1001][1001],br[1001][1001],bc[1001][1001],r[1001][1001],c[1001][1001]; 4 char p[1001][1001]; 5 int main() 6 { 7 int t,i,j,k,n,m,max,min,num = 0; 8 scanf("%d",&t); 9 while(t--) 10 { 11 num ++; 12 memset(rr,0,sizeof(rr)); 13 memset(rc,0,sizeof(rc)); 14 memset(br,0,sizeof(br)); 15 memset(bc,0,sizeof(bc)); 16 memset(r,0,sizeof(r)); 17 memset(c,0,sizeof(c)); 18 scanf("%d%d%*c",&n,&m); 19 for(i = 0; i <= n-1; i ++) 20 gets(p[i]); 21 r[0][0] = 1; 22 c[0][0] = 1; 23 if(p[0][0] == 'R') 24 { 25 rr[0][0] = 1; 26 rc[0][0] = 1; 27 } 28 else 29 { 30 br[0][0] = 1; 31 bc[0][0] = 1; 32 } 33 for(i = 1; i <= n-1; i ++) 34 { 35 if(p[i][0] == 'R') 36 { 37 rr[i][0] = rr[i-1][0]+1; 38 rc[i][0] = 1; 39 } 40 else 41 { 42 rr[i][0] = 0; 43 rc[i][0] = 0; 44 } 45 if(p[i][0] == 'B') 46 { 47 br[i][0] = br[i-1][0]+1; 48 bc[i][0] = 1; 49 } 50 else 51 { 52 br[i][0] = 0; 53 bc[i][0] = 0; 54 } 55 if(p[i][0] != p[i-1][0]) 56 { 57 r[i][0] = r[i-1][0]+1; 58 c[i][0] = 1; 59 } 60 else 61 { 62 r[i][0] = 1; 63 c[i][0] = 1; 64 } 65 } 66 for(i = 1; i <= m-1; i ++) 67 { 68 if(p[0][i] == 'R') 69 { 70 rc[0][i] = rc[0][i-1]+1; 71 rr[0][i] = 1; 72 } 73 else 74 { 75 rc[0][i] = 0; 76 rr[0][i] = 0; 77 } 78 if(p[0][i] == 'B') 79 { 80 bc[0][i] = bc[0][i-1]+1; 81 br[0][i] = 1; 82 } 83 else 84 { 85 bc[0][i] = 0; 86 br[0][i] = 0; 87 } 88 if(p[0][i] != p[0][i-1]) 89 { 90 c[0][i] = c[0][i-1]+1; 91 r[0][i] = 1; 92 } 93 else 94 { 95 c[0][i] = 1; 96 r[0][i] = 1; 97 } 98 } 99 for(i = 1; i <= n-1; i ++) 100 { 101 for(j = 1; j <= m-1; j ++) 102 { 103 if(p[i][j] == 'R') 104 { 105 rr[i][j] = rr[i-1][j]+1; 106 rc[i][j] = rc[i][j-1]+1; 107 br[i][j] = 0; 108 bc[i][j] = 0; 109 } 110 else if(p[i][j] == 'B') 111 { 112 br[i][j] = br[i-1][j]+1; 113 bc[i][j] = bc[i][j-1]+1; 114 rr[i][j] = 0; 115 rc[i][j] = 0; 116 } 117 if(p[i][j] != p[i-1][j-1]) 118 { 119 r[i][j] = 1; 120 c[i][j] = 1; 121 } 122 else 123 { 124 if(p[i][j] != p[i-1][j]&&p[i][j] != p[i][j-1]) 125 { 126 r[i][j] = r[i-1][j]+1; 127 c[i][j] = c[i][j-1]+1; 128 } 129 else if(p[i][j] != p[i-1][j]&&p[i][j] == p[i][j-1]) 130 { 131 r[i][j] = r[i-1][j]+1; 132 c[i][j] = 1; 133 } 134 else if(p[i][j] == p[i-1][j]&&p[i][j] != p[i][j-1]) 135 { 136 r[i][j] = 1; 137 c[i][j] = c[i][j-1]+1; 138 } 139 else 140 { 141 r[i][j] = 1; 142 c[i][j] = 1; 143 } 144 } 145 } 146 } 147 max = 0; 148 for(i = 0; i <= n-1; i ++) 149 { 150 for(j = 0; j <= m-1; j ++) 151 { 152 if(max < 2*(br[i][j]+bc[i][j])) 153 { 154 min = br[i][j]; 155 k = 1; 156 if(max < 2*(k+min)) 157 max = 2*(k+min); 158 for(k = 2; k <= bc[i][j]; k ++) 159 { 160 if(min > br[i][j-k+1]) 161 min = br[i][j-k+1]; 162 if(max < 2*(k+min)) 163 { 164 max = 2*(k+min); 165 } 166 } 167 } 168 if(max < 2*(rr[i][j]+rc[i][j])) 169 { 170 min = rr[i][j]; 171 k = 1; 172 if(max < 2*(k+min)) 173 max = 2*(k+min); 174 for(k = 2; k <= rc[i][j]; k ++) 175 { 176 if(min > rr[i][j-k+1]) 177 min = rr[i][j-k+1]; 178 if(max < 2*(k+min)) 179 { 180 max = 2*(k+min); 181 } 182 } 183 } 184 min = r[i][j]; 185 if(min > c[i][j]) 186 min = c[i][j]; 187 if(max < 4*min) 188 max = 4*min; 189 } 190 } 191 printf("Case #%d: %d\n",num,max); 192 } 193 return 0; 194 }
