hdu 4419(离散化+扫描线)

题意：有三种颜色的矩形，红R绿G蓝B，然后给出每个矩形左下角和右上角坐标，矩阵重叠会形成新的颜色的矩形，颜色有RG、GR、GB、RGB，一共七种颜色，问所有矩形放好后，每种颜色的面积并是多少。

题解：一开始写挫了，RGB分别用1 2 3代表，然后pushup函数写了很长很长，后来发现如果RBG分别用1 2 4代表就能用位运算来写，代码长度顿时少了很多。3是RB也就是1|2,5是RG也就是1|4，6是BG也就是2|4,7是RBG也就是1|2|4。用线段树维护每种颜色在当前区间内所占长度，然后另一个线段树数组维护当前区间都有什么颜色。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
const int N = 20005;
struct Line {
int lx, rx, h;
int v;
Line(int a, int b, int c, int d): lx(a), rx(b), h(c), v(d) {}
bool operator < (const Line& a) const { return h < a.h; }
};
int n, tree[N << 2][8], set[N << 2][4];
long long res[8];
vector<int> a;
vector<Line> line;
map<int, int> mp;

void pushup(int k, int left, int right) {
int flag = 0;
if (set[k][0] > 0) flag |= 1;
if (set[k][1] > 0) flag |= 2;
if (set[k][2] > 0) flag |= 4;
for (int i = 1; i <= 7; i++)
tree[k][i] = 0;
if (flag) {
tree[k][flag] = tree[k][0];
for (int i = 1; i <= 7; i++) {
if (flag != (flag | i)) {
int temp = 0;
if (left + 1 != right)
temp = tree[k * 2][i] + tree[k * 2 + 1][i];
tree[k][flag | i] += temp;
tree[k][flag] -= temp;
}
}
} else if (left + 1 != right) {
for (int i = 1; i <= 7; i++)
tree[k][i] = tree[k * 2][i] + tree[k * 2 + 1][i];
}
}

void build(int k, int left, int right) {
set[k][0] = set[k][1] = set[k][2] = 0;
tree[k][0] = a[right] - a[left];
for (int i = 1; i <= 7; i++)
tree[k][i] = 0;
if (left + 1 != right) {
int mid = (left + right) / 2;
build(k * 2, left, mid);
build(k * 2 + 1, mid, right);
}
}

void modify(int k, int left, int right, int l, int r, int x) {
if (l <= left && right <= r) {
int temp;
if (abs(x) == 1) temp = 0;
else if (abs(x) == 2) temp = 1;
else temp = 2;
set[k][temp] += (x > 0 ? 1 : -1);
pushup(k, left, right);
return;
}
int mid = (left + right) / 2;
if (l < mid)
modify(k * 2, left, mid, l, r, x);
if (r > mid)
modify(k * 2 + 1, mid, right, l, r, x);
pushup(k, left, right);
}

int main() {
int t, cas = 1;
scanf("%d", &t);
while (t--) {
a.clear(), line.clear(), mp.clear();
scanf("%d", &n);
char col[5];
int x1, y1, x2, y2;
for (int i = 0; i < n; i++) {
int temp;
scanf("%s%d%d%d%d", col, &x1, &y1, &x2, &y2);
if (col[0] == 'R') temp = 1;
else if (col[0] == 'G') temp = 2;
else temp = 4;
line.push_back(Line(x1, x2, y1, temp));
line.push_back(Line(x1, x2, y2, -temp));
a.push_back(x1);
a.push_back(x2);
}
sort(line.begin(), line.end());
sort(a.begin(), a.end());
a.erase(unique(a.begin(), a.end()), a.end());
int sz = a.size(), sz2 = line.size();
for (int i = 0; i < sz; i++)
mp[a[i]] = i;
build(1, 0, sz - 1);
memset(res, 0, sizeof(res));
for (int i = 0; i < sz2; i++) {
if (i != 0) {
int temp = line[i].h - line[i - 1].h;
for (int j = 1; j <= 7; j++)
res[j] += (long long)tree[1][j] * temp;
}
modify(1, 0, sz - 1, mp[line[i].lx], mp[line[i].rx], line[i].v);
}
printf("Case %d:\n", cas++);
printf("%lld\n%lld\n%lld\n%lld\n%lld\n%lld\n%lld\n", res[1], res[2], res[4], res[3], res[5], res[6], res[7]);
}
return 0;
}