Edge Detection 解题报告

Edge Detection 解题报告 计算机科学技术系赵静 09908114 zhaojing@db.pku.edu.cn

题型：模拟类 85 85 85 85 85 0 0 85 85 85 85 85 75 75 Input Image Output Image

关键点：时空限制 • 空间限制：sizeof (char) * 1000000000 = 1GB • 逐个计算 • 只保留当前像素及周围像素的值 • 随时输出计算结果 • 时间限制 • 尽量避免重复计算

数据结构 • 输入数据 • long imageWidth; • int pixVal[MAXRUNS + 2]; // pixel values in runs • long runLength[MAXRUNS + 2]; // runs • int lastRunIndex; • 用于计算的关键数据结构 • int runIndex[3]; // index of run containing pixel • long leftInRun[3]; // number of further pixels in the run • long column; // column of pixel in center of filter, 0 to imageWidth - 1

runIndex[0]=1 runIndex[1]=2 runIndex[2]=2 leftInRun[0]=2 leftInRun[1]=10 leftInRun[2]=3 数据结构图示 runLength pixVal

算法（1）：计算单一目的像素值 int edgeFilter() { int mid = pixVal[runIndex[1]]; // pixel value in mid row int max = 0; for (int r = 0; r < 3; r++) // for each row used in the filter if (runIndex[r] > 0 && runIndex[r] <= lastRunIndex) { // skip dummy rows max = updateMax(runIndex[r], mid, max); if (column > 0 && leftInRun[r] == runLength[runIndex[r]] - 1) max = updateMax(runIndex[r]-1, mid, max); if (column < imageWidth -1 && leftInRun[r] == 0) max = updateMax(runIndex[r]+1, mid, max); } return max; }

算法（2）：减少重复计算的策略 leftInRun[0]=2 leftInRun[1]=10 long minRun() { long minAfter = 2000000000; for (int r = 0; r < 3; r++) { if (leftInRun[r] == runLength[runIndex[r]] - 1) return 1; if (minAfter > leftInRun[r]) minAfter = leftInRun[r]; } if (minAfter == 0) return 1; return minAfter; } leftInRun[2]=3 minAfter=2

算法（3）：整体流程 outBufferPixVal = edgeFilter(); outBufferRun = minRun(); advance(outBufferRun); // skip past all the pixels calculated while (runIndex[1] <= lastRunIndex) { outPixVal = edgeFilter(); outRun = minRun(); advance(outRun); if (outPixVal == outBufferPixVal) // combine runs with equal values outBufferRun += outRun; else // output old runLength, save new one { cout << outBufferPixVal << " " << outBufferRun << endl; outBufferPixVal = outPixVal; outBufferRun = outRun; } }

值得注意的细节：起始状态 runLength[0] = 2*imageWidth; // add initial border runLength[lastRunIndex--] = 2*imageWidth; // add end border for (int r = 0; r < 3; r++) { runIndex[r] = 0; // dummy run index before actual image leftInRun[r] = (3-r)*imageWidth - 1; // pixels -3,-2,-1 rows into image } advance(2*imageWidth); // +2 rows to first pixels -1,0,1 rows into image runIndex[0]=0 leftInRun[0]=6 runIndex[1]=1 leftInRun[1]=3 runIndex[2]=2 leftInRun[2]=14

The End Thank you for your attention!

Edge Detection 解题报告