hdu1543 Paint the Wall

表示之前对离散化只是耳闻而已。。。今天看了一下感觉是用映射来实现将一个无限的空间映射到有限空间中去。。然后百度知道此题是用离散化+二分来实现，参照网上代码写了下还是WA。。。还是理解不够深刻，之前会WA居然是因为没有区分好单复数，要同时考虑is和are还有加不加s，表示一直在被英文虐。。。
#include <iostream>
using namespace std;
const int size = 110;
int x[2*size], y[2*size];
int ax[size], bx[size], ay[size], by[size], colo[size];
bool cmp(int a, int b)
{
return a < b;
}
int n;
/*int BinaryS(int ans, int st, int ed, int *t)
{
int left = st, right = ed;
while (left < right){
int mid = (left+right)/2;
if (t[mid] == ans)return mid;
if (t[mid] < ans){
left = mid+1;
}
if (t[mid] > ans){
right = mid-1;
}
}
return 0;
}*/
int BinaryS(int ans, int st, int ed, int *x)
{
int left = st, right = ed;
while (left <= right){
int mid = (left+right)/2;
//cout<<"x:"<<x[mid]<<"ans:"<<ans<<endl;
if (x[mid] == ans){ return mid;}
if (x[mid] < ans){
left = mid+1;
}
if (x[mid] > ans){
right = mid-1;
}
}
//return 0;
}
int main()
{
int w, h;
int nc = 0;
while (scanf("%d%d", &h, &w) && (w+h)){
int nx, ny;
nx = ny = 0;
scanf("%d", &n);
for (int i = 0; i < n; i ++){
scanf("%d%d%d%d%d", &ax[i], &ay[i], &bx[i], &by[i], &colo[i]);
x[nx ++] = ax[i], y[ny ++] = ay[i];
x[nx ++] = bx[i], y[ny ++] = by[i];
}
int xx = 0, yy = 0;
for (int i = 1; i < nx; i ++){
if (x[i] != x[i-1]){
xx ++;
x[xx] = x[i];
}
}
for (int i = 1; i < ny; i ++){
if (y[i] != y[i-1]){
yy ++;
y[yy] = y[i];
}
}
sort(x, x+xx+1, cmp);
sort(y, y+yy+1, cmp);
/*for (int i = 0; i <= xx; i ++){
cout<<x[i]<<' ';
}
cout<<endl<<endl;*/
int map1[2*size][2*size] = {0};
for (int i = 0; i < n; i ++){
//cout<<"ax:"<<ax[i]<<endl;
int sx = BinaryS(ax[i], 0, xx, x);
int sy = BinaryS(ay[i], 0, yy, y);
int ex = BinaryS(bx[i], 0, xx, x);
int ey = BinaryS(by[i], 0, yy, y);
for (int ii = sx; ii < ex; ii ++){
for (int j = sy; j < ey; j ++){
map1[ii][j] = colo[i];
}
}
//cout<<"sx:"<<sx<<' '<<"ex:"<<ex<<endl;
//cout<<"sy:"<<sy<<' '<<"ey:"<<ey<<endl;
}
/*for (int i = 0; i < 10; i ++){
for (int j = 0; j < 10; j ++){
cout<<map1[i][j];
}
cout<<endl;
}*/
int colour[size] = {0};
for (int i = 0; i < xx; i ++){
for (int j = 0; j < yy; j ++){
colour[map1[i][j]] += (x[i+1]-x[i])*(y[j+1]-y[j]);
}
}
if (nc)printf("\n");
nc ++;
printf("Case %d:\n", nc);
int ans = 0;
for (int i = 1; i <= 100; i ++){
if (colour[i]){
ans ++;
printf("%d %d\n", i, colour[i]);
}
}
printf("There are %d color", ans);
if (ans > 1)printf("s");
printf(" left on the wall.\n");
//cout<<"xx:"<<xx<<endl;
}
return 0;
}