POJ 1009.Edge Detection

题目:http://poj.org/problem?id=1009

AC代码(C++):

#include <iostream>#include <algorithm>#include <stdio.h>#include <vector>#include <queue>#include <math.h>using namespace std;struct PAIR{int ccc;int num;};struct POINT{int ccc;int r;int c;};int n;vector<PAIR> input;vector<POINT> needToCa;vector<PAIR> output;long long pointCount;int rowCount;int getC(int r,int c){long long pos = r*n + c;long long startPos = 0;for(vector<PAIR>::iterator it = input.begin(); it != input.end(); it++){if(startPos<=pos&&startPos+it->num>pos)return it->ccc;startPos += it->num;}}int getCCC(int r, int c){int rtn = 0;for(int i=r-1;i<=r+1;i++){        for(int j=c-1;j<=c+1;j++)        {            if(i<0||i>=rowCount||j>=n||j<0||(i==r&&j==c))continue;            int temp = abs(getC(i,j)-getC(r,c));            if(rtn<temp)rtn=temp;        }}return rtn;}bool cmp(POINT a, POINT b){if(a.r!=b.r)return a.r<b.r;else return a.c<b.c;}int main(){for(;;){cin>>n;if(n==0)break;//输入for(;;){PAIR tmppair;cin>>tmppair.ccc>>tmppair.num;if(tmppair.ccc!=0||tmppair.num!=0){input.push_back(tmppair);pointCount+=tmppair.num;}else{input.push_back(tmppair);//这是个坑break;}}rowCount = pointCount/n;//计算关键像素点long long startPos = 0;for(vector<PAIR>::iterator it = input.begin(); it != input.end(); it++){int tmpr;int tmpc;tmpr = startPos/n;tmpc = startPos%n;for(int i = tmpr-1; i <= tmpr+1; i++){for(int j = tmpc-1; j <= tmpc+1; j++){if(i>=0&&i<rowCount&&j>=0&&j<n){POINT tmppoint;tmppoint.r = i;tmppoint.c = j;tmppoint.ccc = getCCC(i,j);needToCa.push_back(tmppoint);}}}startPos += it->num;}//排序sort(needToCa.begin(),needToCa.end(),cmp);//输出 cout<<n<<endl;vector<POINT>::iterator temp = needToCa.begin();        for(vector<POINT>::iterator it = needToCa.begin(); it != needToCa.end(); it++)        {            if(it->ccc==temp->ccc)                continue;            printf("%d %d\n",temp->ccc,it->r*n+it->c-temp->r*n-temp->c);            temp=it;        }        printf("%d %d\n",temp->ccc,pointCount-temp->r*n-temp->c);        cout<<"0 0\n";//清理input.clear();needToCa.clear();output.clear();pointCount = 0;}cout<<"0\n";}

总结:

1.像如此大规模的模拟题暴力肯定不能解决, 需要找规律. 而这题的规律是每个颜色-数量对(PAIR)的起始像素和它周围的8个像素可能是输出PAIR的起始位置. 也就是说, 只要计算所有输入PAIR的起始像素及其周围像素(即关键像素)的输出颜色, 除此之外的像素颜色就等于它们前面最近的关键像素点的颜色.

2.最后输入的0 0不能忽略, 也要放进去计算. 具体看下面这个例子

3

10 8

0 1

0 0