poj 1151 Atlantis

类型：离散化

题目：http://poj.org/problem?id=1151

来源：Mid-Central European Regional Contest 2000

思路【一】：

（1）使用map离散化

（2）对每个矩形进行覆盖操作

（3）统计覆盖的区域面积和

//poj 1151 Atlantis
// wa ac 312K 32MS
#include <iostream>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;const int MAXM = 2610;
int n, cntx, cnty, cas = 1;
bool endd[210][210];
double dax[MAXM], day[MAXM];
map<double, int> mpx, mpy, mpxx, mpyy;
map<double, int>::iterator it;
struct pot {double x1, y1, x2, y2;
}p[MAXM];void solve() {int i, j, k;cntx = cnty = 1;for(it = mpxx.begin(); it != mpxx.end(); ++it)mpx[it->first] = cntx, dax[cntx++] = it->first;for(it = mpyy.begin(); it != mpyy.end(); ++it)mpy[it->first] = cnty, day[cnty++] = it->first;for(i = 1; i <= n; ++i) {int x1 = mpx[p[i].x1];int y1 = mpy[p[i].y1];int x2 = mpx[p[i].x2];int y2 = mpy[p[i].y2];for(j = x1; j < x2; ++j)for(k = y1; k < y2; ++k)endd[j][k] = true;}double sum = 0.0;for(i = 1; i < cntx - 1; ++i)for(j = 1; j < cnty - 1; ++j)if(endd[i][j]) {sum += (dax[i + 1] - dax[i]) * (day[j + 1] - day[j]);}printf("Test case #%d\n", cas++);printf("Total explored area: %.2lf\n\n", sum);
}int main() {double x1, x2, y1, y2;int i;while(scanf("%d", &n) != EOF, n) {cntx = cnty = 1;mpx.clear();mpy.clear();mpxx.clear();mpyy.clear();memset(endd, false, sizeof(endd));for(i = 1; i <= n; ++i) {scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);p[i].x1 = x1, p[i].x2 = x2, p[i].y1 = y1, p[i].y2 = y2;if(mpxx.count(x1) == 0)mpxx[x1] = cntx;if(mpyy.count(y1) == 0)mpyy[y1] = cnty;if(mpxx.count(x2) == 0)mpxx[x2] = cntx;if(mpyy.count(y2) == 0)mpyy[y2] = cnty;}solve();}return 0;
}

思路【二】：

（1）对y坐标离散化，对x坐标建立扫描线

（2）以y坐标建立线段树

（3）插入线段【x扫描线】，每插入一次，统计可得到的矩形面积

（4）对于每条线段，如果有p[t].c > 0 即可统计，当p[t].c = 0时该块矩形扫描完成，不再进行统计

//poj 1151 Atlantis
//ac 224K 0MS
#include <iostream>
#include <sstream>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;#define FOR(i,a,b) for(i = (a); i < (b); ++i)
#define FORE(i,a,b) for(i = (a); i <= (b); ++i)
#define FORD(i,a,b) for(i = (a); i > (b); --i)
#define FORDE(i,a,b) for(i = (a); i >= (b); --i)const int MAXN = 1010;int n, m, len, t = 1;
struct node {int l, r, c;double len, sum;
}p[MAXN * 5];
struct line {double x, y1, y2;int sign;
}l[MAXN];
double Y[MAXN];void creat(int t, int l, int r) {int mid = (l + r) >> 1;p[t].l = l, p[t].r = r;p[t].c = 0, p[t].sum = 0;p[t].len = Y[r] - Y[l];if(l + 1 < r) {creat(t * 2, l, mid);creat(t * 2 + 1, mid, r);return;}
}void insert(int t, int l, int  r, int flag) {int mid = (p[t].l + p[t].r) / 2;if(p[t].l == l && p[t].r == r) {(flag == true) ? ++p[t].c : --p[t].c;}else {if(r <= mid)insert(t * 2, l, r, flag);else if(l >= mid)insert(t * 2 + 1, l, r, flag);else {insert(t * 2, l, mid, flag);insert(t * 2 + 1, mid, r, flag);}}if(p[t].c > 0)p[t].sum = p[t].len;else if(p[t].r - p[t].l > 1)p[t].sum = p[t * 2].sum + p[t * 2 + 1].sum;else // !!!p[t].sum = 0;
}int cmp(line a, line b) {return a.x < b.x;
}int find(double key) {int l = 0, r = len - 1, mid;while(l <= r) {mid = (l + r) >> 1;if(Y[mid] >= key)r = mid - 1;elsel = mid + 1;}return l;
}void solve() {int i;double x1, x2, y1, y2, ans = 0;m = 0;FORE(i, 1, n) {scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);Y[m] = y1; l[m].x = x1; l[m].y1 = y1; l[m].y2 = y2; l[m++].sign = 1;Y[m] = y2; l[m].x = x2; l[m].y1 = y1; l[m].y2 = y2; l[m++].sign = 0;}sort(Y, Y + m);sort(l, l + m, cmp);len = unique(Y, Y + m) - Y;creat(1, 0, len - 1);FOR(i, 0, m) {if(l[i].sign)insert(1, find(l[i].y1), find(l[i].y2), 1);elseinsert(1, find(l[i].y1), find(l[i].y2), 0);ans += p[1].sum * (l[i + 1].x - l[i].x);}printf("Test case #%d\n", t++);printf("Total explored area: %.2lf\n\n", ans);
}int main() {while(scanf("%d", &n) != EOF, n) {solve();}return 0;
}
/*
2
10 10 20 20
15 15 25 25.5
0
*/

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