uva 1526 - Edge Detection（二分+排序）

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
[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;
else
r = 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;
}