uva 1526 - Edge Detection（二分+排序）

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题目链接：uva 1526 - Edge Detection

题目大意：先给出width，表示说一个w*w的图片，每个位置上的数值代表该位置的像素，现在有一种算法用于装换图片，转换后图片的表示方式为，每个位置上的数值为原先图上位置的像素与周围8个位置像素之差的绝对值的最大值。

然后图的表示给出和要求输出方式为run length encoding，即由若干对两位数组成，分别表示像素和连续个数。给出原图的RLE，输出转换后的RLE。

解题思路：首先会受到影响的肯定为交界处，所以找出所有的交界处，将周围8个数取出单独考虑，在查找值时用二分。

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>using namespace std;const int N = 1005;int inN, outN, inData[N][2];int width, total;struct state {int val, pos;}outData[N*9];void init () {total = inN = outN = 0;memset(inData, 0, sizeof(inData));int a, b;while (scanf("%d%d", &a, &b) == 2) {inData[inN][0] = a;inData[inN][1] = total;total += b;if (0 == b) break;++inN;}}int getValue (int pos) {int l = 0, r = inN - 1;while (l <= r) {int mid = (l + r) / 2;if (inData[mid][1] <= pos)l = mid + 1;elser = mid - 1;}return inData[r][0];}int cal(int pos) {int r = pos / width;int l = pos % width;int ans = 0;for (int i = r - 1; i <= r + 1; i++) {for (int j = l - 1; j <= l + 1; j++) {int p = i * width + j;if (i < 0 || j < 0 || j >= width || p >= total || p == pos)continue;int t = abs(getValue(p) - getValue(pos));ans = max(ans, t);}}return ans;}inline bool cmp (const state& a, const state& b) {return a.pos < b.pos;}void solve () {for (int i = 0; i <= inN; i++) {int r = inData[i][1] / width;int l = inData[i][1] % width;for (int j = r - 1; j <= r + 1; j++) {for (int k = l - 1; k <= l + 1; k++) {int p = j * width + k;if (j < 0 || k < 0 || p >= total || p < 0)continue;outData[outN].val = cal(p);outData[outN].pos = p;++outN;}}}sort (outData, outData + outN, cmp);state cur = outData[0];for (int i = 0; i < outN; i++) {if (cur.val == outData[i].val) continue;printf("%d %d\n", cur.val, outData[i].pos - cur.pos);cur = outData[i];}printf("%d %d\n", cur.val, total - cur.pos);printf("0 0\n");}int main () {while (scanf("%d", &width) == 1 && width) {init ();printf("%d\n", width);solve ();}printf("0\n");return 0;}

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