标签:style http color io os ar for sp on
题目链接:hdu 3642 Get The Treasury
题目大意:三维坐标系,给定若干的长方体,问说有多少位置被覆盖3次以上。
解题思路:扫描线,将第三维分离出来,就是普通的二维扫描线,然后对于每个节点要维护覆盖0,1,2,3以上这4种的覆盖面积。
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 4005;
vector<int> pos;
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], v[maxn << 2], s[maxn << 2][5];
inline void pushup(int u) {
memset(s[u], 0, sizeof(s[u]));
if (v[u] >= 3)
s[u][3] = pos[rc[u]+1] - pos[lc[u]];
else {
if (lc[u] == rc[u])
s[u][v[u]] = pos[rc[u]+1] - pos[lc[u]];
else if (v[u] == 2) {
s[u][2] = s[lson(u)][0] + s[rson(u)][0];
for (int i = 1; i <= 3; i++)
s[u][3] += s[lson(u)][i] + s[rson(u)][i];
} else if (v[u] == 1) {
s[u][1] = s[lson(u)][0] + s[rson(u)][0];
s[u][2] = s[lson(u)][1] + s[rson(u)][1];
for (int i = 2; i <= 3; i++)
s[u][3] += s[lson(u)][i] + s[rson(u)][i];
} else {
for (int i = 0; i <= 3; i++)
s[u][i] = s[lson(u)][i] + s[rson(u)][i];
}
}
}
inline void maintain(int u, int d) {
v[u] += d;
pushup(u);
}
void build (int u, int l, int r) {
lc[u] = l;
rc[u] = r;
v[u] = 0;
if (l == r) {
maintain(u, 0);
return ;
}
int mid = (l + r) / 2;
build (lson(u), l, mid);
build (rson(u), mid + 1, r);
pushup(u);
}
void modify (int u, int l, int r, int d) {
if (l <= lc[u] && rc[u] <= r) {
maintain(u, d);
return;
}
int mid = (lc[u] + rc[u]) / 2;
if (l <= mid)
modify(lson(u), l, r, d);
if (r > mid)
modify(rson(u), l, r, d);
pushup(u);
}
struct Seg {
int x, l, r, d;
Seg (int x = 0, int l = 0, int r = 0, int d = 0) {
this->x = x;
this->l = l;
this->r = r;
this->d = d;
}
};
typedef long long ll;
vector<Seg> g[1005];
inline bool cmp (const Seg& a, const Seg& b) {
return a.x < b.x;
}
void init () {
int n, x1, x2, y1, y2, z1, z2;
scanf("%d", &n);
for (int i = 0; i <= 1000; i++)
g[i].clear();
for (int i = 0; i < n; i++) {
scanf("%d%d%d%d%d%d", &x1, &y1, &z1, &x2, &y2, &z2);
for (int i = z1; i < z2; i++) {
g[i + 500].push_back(Seg(x1, y1, y2, 1));
g[i + 500].push_back(Seg(x2, y1, y2, -1));
}
}
}
inline int find (int k) {
return lower_bound(pos.begin(), pos.end(), k) - pos.begin();
}
ll solve (int idx) {
if (g[idx].size() == 0)
return 0;
ll ret = 0;
pos.clear();
sort(g[idx].begin(), g[idx].end(), cmp);
for (int i = 0; i < g[idx].size(); i++) {
pos.push_back(g[idx][i].l);
pos.push_back(g[idx][i].r);
}
sort(pos.begin(), pos.end());
build(1, 0, pos.size());
for (int i = 0; i < g[idx].size(); i++) {
modify(1, find(g[idx][i].l), find(g[idx][i].r) - 1, g[idx][i].d);
if (i + 1 != g[idx].size())
ret += 1LL * s[1][3] * (g[idx][i+1].x - g[idx][i].x);
}
return ret;
}
int main () {
int cas;
scanf("%d", &cas);
for (int kcas = 1; kcas <= cas; kcas++) {
init();
ll ans = 0;
for (int i = 0; i <= 1000; i++)
ans += solve(i);
printf("Case %d: %I64d\n", kcas, ans);
}
return 0;
}
hdu 3642 Get The Treasury(扫描线)
标签:style http color io os ar for sp on
原文地址:http://blog.csdn.net/keshuai19940722/article/details/39997151