http://acm.hdu.edu.cn/showproblem.php?pid=1542
我的做法是把x轴的表示为线段，然后更新y
不考虑什么优化的话，开始的时候，把他们表达成线段，并按y排序，然后第一次加入线段树的应该就是最底下那条，然后第二条的时候，我们可以询问第二条那段区间，有多少是已经被覆盖的，然后把面积算上就可以。
所以如果区间都是整数，而且数值很少，那么就是线段树成段覆盖的问题了。但是这里是浮点数而且很大。
所以只能把它离散化。
这个时候线段树就不是连续的了，这里就有bug，问题就变成了怎么表达这颗线段树了。
思路是把它弄成L + 1 == R就是叶子节点，这样的话，每个节点都保存了一个区间了，
例如
要保存5、10、15、30
一般的线段树
5、30
5、10 15、30
5 10 15 30
但是这样怎么表示[10, 15]这段区间呢？
所以把线段树变成
5、30
5、10 10、 30
10、15 15、30
就行了。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
int n;
const int maxn = 2e2 + 20;
struct node {
int L, R;
double x1, x2, y;
int cover;
} seg[maxn << 2];
struct info {
double x1, x2, y;
int flag;
bool operator < (const struct info & rhs) const {
return y < rhs.y;
}
} in[maxn];
double xx[maxn];
void build(int L, int R, int cur) {
seg[cur].cover = 0;
seg[cur].x1 = xx[L];
seg[cur].x2 = xx[R];
seg[cur].y = -1;
seg[cur].L = L;
seg[cur].R = R;
if (L + 1 == R) { //两个节点作为一个叶子
return;
}
int mid = (L + R) >> 1;
build(L, mid, cur << 1);
build(mid, R, cur << 1 | 1);
}
double upDate(int begin, int end, double y, int flag, int cur) {
if (end <= seg[cur].L || begin >= seg[cur].R) return 0;
if (seg[cur].L + 1 == seg[cur].R) {
if (seg[cur].cover) {
double ans = (seg[cur].x2 - seg[cur].x1) * (y - seg[cur].y);
seg[cur].cover += flag;
seg[cur].y = y;
return ans;
} else {
seg[cur].cover += flag;
seg[cur].y = y;
return 0;
}
}
double ans = upDate(begin, end, y, flag, cur << 1) + upDate(begin, end, y, flag, cur << 1 | 1);
return ans;
}
void work() {
int lenin = 0;
int lenxx = 0;
for (int i = 1; i <= n; ++i) {
double xx1, xx2, yy1, yy2;
scanf("%lf%lf%lf%lf", &xx1, &yy1, &xx2, &yy2);
lenin++;
in[lenin].x1 = xx1;
in[lenin].x2 = xx2;
in[lenin].y = yy1;
in[lenin].flag = 1;
lenxx++;
xx[lenxx] = xx1;
lenin++;
in[lenin].x1 = xx1;
in[lenin].x2 = xx2;
in[lenin].y = yy2;
in[lenin].flag = -1;
lenxx++;
xx[lenxx] = xx2;
}
sort(in + 1, in + 1 + lenin);
sort(xx + 1, xx + 1 + lenxx);
lenxx = unique(xx + 1, xx + 1 + lenxx) - (xx + 1);
build(1, lenxx, 1);
double ans = 0;
const int root = 1;
for (int i = 1; i <= lenxx; ++i) {
printf("%f**\n", xx[i]);
}
for (int i = 1; i <= lenin; ++i) {
int L = lower_bound(xx + 1, xx + 1 + lenxx, in[i].x1) - xx;
int R = lower_bound(xx + 1, xx + 1 + lenxx, in[i].x2) - xx;
// cout << L << " " << R << endl;
ans += upDate(L, R, in[i].y, in[i].flag, root);
}
static int f = 0;
printf("Test case #%d\n", ++f);
printf("Total explored area: %0.2f\n", ans);
// printf("%0.2f\n", ans);
}
int main() {
#ifdef local
freopen("data.txt","r",stdin);
#endif
while (scanf("%d", &n) != EOF && n) {
work();
printf("\n");
}
return 0;
}
View Code
#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int maxn = 1e3 + 20;
struct Node {
double x1, x2, y;
int L, R;
int cover;
}seg[maxn << 2];
vector<double> vc;
int n;
struct Info {
double x1, x2, y;
int cover;
bool operator < (const struct Info & rhs) const {
return y < rhs.y;
}
}fuck[maxn * 2];
void build(int L, int R, int cur) {
seg[cur].L = L, seg[cur].R = R;
seg[cur].x1 = vc[L], seg[cur].x2 = vc[R];
seg[cur].y = -inf;
seg[cur].cover = 0;
if (L + 1 == R) return;
int mid = (L + R) >> 1;
build(L, mid, cur << 1);
build(mid, R, cur << 1 | 1);
}
double add(int x1, int x2, double y, int cover, int cur) {
if (x2 <= seg[cur].L || x1 >= seg[cur].R) return 0;
if (seg[cur].L + 1 == seg[cur].R) {
if (seg[cur].cover) {
double res = (y - seg[cur].y) * (seg[cur].x2 - seg[cur].x1);
seg[cur].cover += cover;
seg[cur].y = y; // 更新最大值y
return res;
} else {
seg[cur].cover += cover;
seg[cur].y = y;
return 0;
}
}
return add(x1, x2, y, cover, cur << 1) + add(x1, x2, y, cover, cur << 1 | 1);
}
void work() {
vc.clear();
vc.push_back(-1.0);
int len = 0;
for (int i = 1; i <= n; ++i) {
double x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
++len;
fuck[len].x1 = x1, fuck[len].x2 = x2, fuck[len].y = y1, fuck[len].cover = 1;
++len;
fuck[len].x1 = x1, fuck[len].x2 = x2, fuck[len].y = y2, fuck[len].cover = -1;
vc.push_back(x1);
vc.push_back(x2);
}
sort(vc.begin(), vc.end());
sort(fuck + 1, fuck + 1 + len);
build(1, vc.size(), 1);
double ans = 0;
for (int i = 1; i <= len; ++i) {
int x1 = lower_bound(vc.begin(), vc.end(), fuck[i].x1) - vc.begin();
int x2 = lower_bound(vc.begin(), vc.end(), fuck[i].x2) - vc.begin();
ans += add(x1, x2, fuck[i].y, fuck[i].cover, 1);
}
static int f = 0;
printf("Test case #%d\n", ++f);
printf("Total explored area: %.2f\n\n", ans);
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
while (cin >> n && n) work();
return 0;
}
View Code
跪着看这篇blog想的
2017年8月15日 13:35:12
http://acm.hdu.edu.cn/showproblem.php?pid=1255
只需要cover >= 2才计算
#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int maxn = 1e3 + 20;
struct Node {
double x1, x2, y;
int L, R;
int cover;
}seg[maxn << 2];
vector<double> vc;
int n;
struct Info {
double x1, x2, y;
int cover;
bool operator < (const struct Info & rhs) const {
return y < rhs.y;
}
}fuck[maxn * 2];
void build(int L, int R, int cur) {
seg[cur].L = L, seg[cur].R = R;
seg[cur].x1 = vc[L], seg[cur].x2 = vc[R];
seg[cur].y = -inf;
seg[cur].cover = 0;
if (L + 1 == R) return;
int mid = (L + R) >> 1;
build(L, mid, cur << 1);
build(mid, R, cur << 1 | 1);
}
double add(int x1, int x2, double y, int cover, int cur) {
if (x2 <= seg[cur].L || x1 >= seg[cur].R) return 0;
if (seg[cur].L + 1 == seg[cur].R) {
if (seg[cur].cover >= 2) {
double res = (y - seg[cur].y) * (seg[cur].x2 - seg[cur].x1);
seg[cur].cover += cover;
seg[cur].y = y; // 更新最大值y
return res;
} else {
seg[cur].cover += cover;
seg[cur].y = y;
return 0;
}
}
return add(x1, x2, y, cover, cur << 1) + add(x1, x2, y, cover, cur << 1 | 1);
}
void work() {
vc.clear();
vc.push_back(-1.0);
int len = 0;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
double x1, y1, x2, y2;
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
++len;
fuck[len].x1 = x1, fuck[len].x2 = x2, fuck[len].y = y1, fuck[len].cover = 1;
++len;
fuck[len].x1 = x1, fuck[len].x2 = x2, fuck[len].y = y2, fuck[len].cover = -1;
vc.push_back(x1);
vc.push_back(x2);
}
sort(vc.begin(), vc.end());
sort(fuck + 1, fuck + 1 + len);
build(1, vc.size(), 1);
double ans = 0;
for (int i = 1; i <= len; ++i) {
int x1 = lower_bound(vc.begin(), vc.end(), fuck[i].x1) - vc.begin();
int x2 = lower_bound(vc.begin(), vc.end(), fuck[i].x2) - vc.begin();
ans += add(x1, x2, fuck[i].y, fuck[i].cover, 1);
}
printf("%.2f\n", ans);
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
int t;
scanf("%d", &t);
while (t--) work();
return 0;
}
View Code
http://codeforces.com/contest/610/problem/D
这题要用到另一种方法
不然TLE
这题是把线条表达成一个长度大小是1的矩形
如果矩形表达成[x1, x2]，那么边长应该是x2 - x1 + 1（因为在直线的意义下所有点都覆盖了）
所以要把矩形表达成[x1 - 1, x2]，同时要注意统一化，就是水平的和垂直的都是按照这个方向改。
x轴要固定向左端减小。
#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int maxn = 400000 + 20;
struct Info {
int x1, x2, cover;
LL y;
bool operator < (const struct Info & rhs) const {
return y < rhs.y;
}
} fuck[maxn];
vector<int> vc;
struct Node {
int L, R;
int x1, x2, y;
int cover;
}seg[maxn << 2];
void build(int L, int R, int cur) {
seg[cur].L = L, seg[cur].R = R;
seg[cur].x1 = vc[L], seg[cur].x2 = vc[R];
seg[cur].y = 0;
seg[cur].cover = 0;
if (L + 1 == R) return;
int mid = (L + R) >> 1;
build(L, mid, cur << 1);
build(mid, R, cur << 1 | 1);
}
void pushUp(int cur) {
if (seg[cur].cover) {
seg[cur].y = vc[seg[cur].R] - vc[seg[cur].L];
} else if (seg[cur].L + 1 == seg[cur].R) {
seg[cur].y = 0;
} else {
seg[cur].y = seg[cur << 1].y + seg[cur << 1 | 1].y;
}
}
void add(int be, int en, int y, int cover, int cur) {
if (seg[cur].L > en || seg[cur].R < be) return;
if (seg[cur].L >= be && seg[cur].R <= en) {
seg[cur].cover += cover;
pushUp(cur);
return;
}
add(be, en, y, cover, cur << 1);
add(be, en, y, cover, cur << 1 | 1);
pushUp(cur);
}
void work() {
vc.clear();
vc.push_back(-inf);
int n, len = 0;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if (x1 == x2) {
if (y1 > y2) swap(y1, y2);
++len;
fuck[len].x1 = x1 - 1, fuck[len].x2 = x2, fuck[len].y = y1 - 1, fuck[len].cover = 1;
++len;
fuck[len].x1 = x1 - 1, fuck[len].x2 = x2, fuck[len].y = y2, fuck[len].cover = -1;
vc.push_back(x1 - 1);
vc.push_back(x2);
} else {
if (x1 > x2) swap(x1, x2);
++len;
fuck[len].x1 = x1 - 1, fuck[len].x2 = x2, fuck[len].y = y1, fuck[len].cover = -1;
++len;
fuck[len].x1 = x1 - 1, fuck[len].x2 = x2, fuck[len].y = y1 - 1, fuck[len].cover = 1;
vc.push_back(x1 - 1);
vc.push_back(x2);
}
}
sort(vc.begin(), vc.end());
// vc.erase(unique(vc.begin(), vc.end()), vc.begin());
sort(fuck + 1, fuck + 1 + len);
build(1, vc.size(), 1);
LL ans = 0;
for (int i = 1; i < len; ++i) {
int x1 = lower_bound(vc.begin(), vc.end(), fuck[i].x1) - vc.begin();
int x2 = lower_bound(vc.begin(), vc.end(), fuck[i].x2) - vc.begin();
add(x1, x2, fuck[i].y, fuck[i].cover, 1);
ans += seg[1].y * (fuck[i + 1].y - fuck[i].y);
}
cout << ans << endl;
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
work();
return 0;
}
View Code
求解区间
[L1, R1]、[L2, R2].....这样区间的，线段树的叶子节点一定要维护两个值。才算是叶子节点。就是上面所说的
主要是：因为这些区间不是连续的。
例如
要保存5、10、15、30
一般的线段树
5、30
5、10 15、30
5 10 15 30
是无法得到a[4] - a[1] = 25的区间长度的。
如果你直接跑线段树，就是左右儿子的总和加上来。这样就是10 - 5 + 30 - 15 = 20
漏了一段，那一段？10--15
所以表示成
5 10 15 30
5 10 10 15 30
10 15 15 30
是一种好的选择，因为这和一般的线段树不同，一般的线段树都是连续的整数，现在是分散的。所以要这样来代表
#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#define lson L, mid, cur << 1
#define rson mid, R, cur << 1 | 1
#define root 1, all, 1
const int maxn = 600000 + 20;
struct Info {
int x1, x2, cover;
LL y;
bool operator < (const struct Info & rhs) const {
return y < rhs.y;
}
} fuck[maxn];
vector<int> vc;
LL seg[maxn << 2], cov[maxn << 2];
void pushUp(int cur, int L, int R) {
if (cov[cur]) seg[cur] = vc[R] - vc[L];
else if (L + 1 == R) seg[cur] = 0;
else {
seg[cur] = seg[cur << 1] + seg[cur << 1 | 1];
}
}
void upDate(int be, int en, int val, int L, int R, int cur) {
if (L > en || R < be) return;
if (L >= be && R <= en) {
cov[cur] += val;
pushUp(cur, L, R);
return;
}
if (L + 1 == R) return; //必须的
int mid = (L + R) >> 1;
upDate(be, en, val, lson);
upDate(be, en, val, rson); //维护两个节点
pushUp(cur, L, R);
}
void work() {
vc.clear();
vc.push_back(-inf);
int n, len = 0;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if (x1 == x2) {
if (y1 > y2) swap(y1, y2);
++len;
fuck[len].x1 = x1 - 1, fuck[len].x2 = x2, fuck[len].y = y1 - 1, fuck[len].cover = 1;
++len;
fuck[len].x1 = x1 - 1, fuck[len].x2 = x2, fuck[len].y = y2, fuck[len].cover = -1;
vc.push_back(x1 - 1);
vc.push_back(x2);
} else {
if (x1 > x2) swap(x1, x2);
++len;
fuck[len].x1 = x1 - 1, fuck[len].x2 = x2, fuck[len].y = y1, fuck[len].cover = -1;
++len;
fuck[len].x1 = x1 - 1, fuck[len].x2 = x2, fuck[len].y = y1 - 1, fuck[len].cover = 1;
vc.push_back(x1 - 1);
vc.push_back(x2);
}
}
// for (int i = 1; i <= n; ++i) {
// int x1, y1, x2, y2;
// cin >> x1 >> y1 >> x2 >> y2;
// vc.push_back(x1);
// vc.push_back(x2);
// ++len;
// fuck[len].x1 = x1, fuck[len].x2 = x2, fuck[len].y = y1, fuck[len].cover = 1;
// ++len;
// fuck[len].x1 = x1, fuck[len].x2 = x2, fuck[len].y = y2, fuck[len].cover = -1;
// }
//上面注释只是普通的矩形面积交。
sort(vc.begin(), vc.end());
sort(fuck + 1, fuck + 1 + len);
int all = vc.size() - 1; //只能去到-1
LL ans = 0;
for (int i = 1; i < len; ++i) {
int x1 = lower_bound(vc.begin(), vc.end(), fuck[i].x1) - vc.begin();
int x2 = lower_bound(vc.begin(), vc.end(), fuck[i].x2) - vc.begin();
upDate(x1, x2, fuck[i].cover, root);
// cout << seg[1] << endl;
ans += seg[1] * (fuck[i + 1].y - fuck[i].y);
}
cout << ans << endl;
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
work();
return 0;
}
矩形面积交数据
2
1 0 3 3
2 1 4 4
