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题目大意:给出n*m的点,可以在上用不同颜色的笔画矩形,菱形,等腰三角形和圆形,因为是一个一个画的,所以有的点会被覆盖掉,原先的颜色就会被覆盖掉了。现在给出每个人画的图案和顺序,问最后每种颜色占了多少个点
解题思路:如果直接暴力的话就会TLE
为了防止被覆盖,就倒着画,如果该点被占有了,就不可以再画了
我们用并查集将每一个点所能到达的最右端的点纪录下来,将那些被使用过的点并起来,然后依次从上往下扫
参考了学长的代码。。。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<cmath>
#include<cstdlib>
#define maxn 210
#define maxm 50010
using namespace std;
int w[maxn][maxm], N, M, q;
int color[10];
map<char, int> Map;
struct Shape{
int x, y, r, w, c, m, Start, End;
void init() {
if(m < 2) {
Start = max(0,x - r);
End = min(N - 1, x + r);
}
else{
r = (r + 1) / 2 - 1;
Start = x;
End = min(N - 1, r + x);
}
}
}S[maxm];
int find(int *f, int x) {
return x == f[x] ? x : f[x] = find(f, f[x]);
}
void init() {
for(int i = 0; i < N; i++)
for(int j = 0; j <= M; j++)
w[i][j] = j;
for(int i = 0; i < 10; i++)
color[i] = 0;
char str[20];
for(int i = 0; i < q; i++) {
scanf("%s", str);
S[i].m = Map[str[0]];
if(str[0] != ‘R‘) {
scanf("%d%d%d%d",&S[i].x, &S[i].y, &S[i].r, &S[i].c);
S[i].init();
}
else{
scanf("%d%d%d%d%d", &S[i].x, &S[i].y, &S[i].r, &S[i].w, &S[i].c);
}
}
}
int count_Rec(int x, int y, int r, int W) {
int cnt = 0;
for(int i = x; i <= x + r - 1 && i < N; i++) {
int t = y;
while(t = find(w[i],t), abs(t - y) < W && t < M) {
cnt++;
w[i][t] = t + 1;
}
}
return cnt;
}
int get_R(int x, int i, int r, int m) {
if(m == 0)
return (int)sqrt(1.0 * r * r - 1.0 * (x - i) * (x - i));
else if(m == 1)
return r - abs(x - i);
else if(m == 2 )
return r - (i - x);
return 0;
}
int count_Oth(int x, int y, int r, int m, int Start, int End) {
int cnt = 0;
for(int i = Start; i <= End; i++) {
int R = get_R(x, i, r, m);
int t = max(0, y - R);
while(t = find(w[i], t) , abs(t - y) <= R && t < M) {
cnt++;
w[i][t] = t + 1;
}
}
return cnt;
}
void solve() {
for(int i = q - 1; i >= 0; i--) {
int &col = color[S[i].c];
if(S[i].m == 3)
col += count_Rec(S[i].x, S[i].y, S[i].r, S[i].w);
else
col += count_Oth(S[i].x, S[i].y, S[i].r, S[i].m,S[i].Start, S[i].End);
}
printf("%d", color[1]);
for(int i = 2; i < 10; i++)
printf(" %d", color[i]);
printf("\n");
}
void begin() {
Map[‘C‘] = 0;
Map[‘D‘] = 1;
Map[‘T‘] = 2;
Map[‘R‘] = 3;
}
int main() {
begin();
while(scanf("%d%d%d", &N, &M, &q) != EOF) {
init();
solve();
}
return 0;
}
UVA - 1493 Draw a Mess 并查集+压缩图
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原文地址:http://blog.csdn.net/l123012013048/article/details/45568839