[UVA][2SAT、SCC、線段交] 11930 - Rectangles
Rectangles
Given a set of rectangles in the plane, determine if it is possible to
choose one diagonal from each rectangle such that none of the
selected diagonals intersect. Two diagonals intersect if they share at
least one point. Note that the rectangles themselves are free
to intersect.
| Rectangles |
Input
Input consists of several test cases. Each test case begins with an integer n (
1
n1000), representing the number of rectangles.
This is followed by n lines each describing a rectangle using 8 integer numbers
x1, y1, x2, y2, x3, y3, x4, y4,
where each
(xi, yi) is a
vertex. All coordinate values are between -109 and 109. The input is terminated by a line containing `0' which
should not be processed.
Output
For each test case, output a line containing either `YES' if the selection is possible or `NO' if not.
Sample Input
4 0 1 1 1 1 0 0 0 1 1 2 1 2 0 1 0 2 3 5 3 5 0 2 0 2 3 3 3 3 2 2 2 7 0 10 10 10 10 0 0 0 10 10 20 10 20 0 10 0 20 30 50 30 50 0 20 0 20 30 30 30 30 20 20 20 30 10 40 10 40 0 30 0 5 0 30 0 30 -10 5 -10 0 0 5 0 5 -10 0 -10 0
Sample Output
YES NO
把對角線兩條分別當作是 val 與 !val
把相交的部分當作推論。
#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <vector>
#include <string.h>
using namespace std;
#define eps 1e-6
struct Pt {
float x, y;
bool operator<(const Pt &a) const {
if(a.x != x)
return x < a.x;
return y < a.y;
}
};
struct Seg {
Pt s, t;
};
float cross(Pt o, Pt a, Pt b) {
return (a.x-o.x)*(b.y-o.y)-(a.y-o.y)*(b.x-o.x);
}
void convexHull(int n, Pt p[], Pt ch[]) {
sort(p, p+n);
int m = 0, i, t;
for(i = 0; i < n; i++) {
while(m >= 2 && cross(ch[m-2], ch[m-1], p[i]) <= 0)
m--;
ch[m++] = p[i];
}
for(i = n-1, t = m+1; i >= 0; i--) {
while(m >= t && cross(ch[m-2], ch[m-1], p[i]) <= 0)
m--;
ch[m++] = p[i];
}
}
int onLine(Seg a, Pt p) {
if((a.s.x-a.t.x)*(p.y-a.s.y) != (a.s.y-a.t.y)*(p.x-a.s.x))
return 0;
if(p.x < a.s.x && p.x < a.t.x)
return 0;
if(p.x > a.s.x && p.x > a.t.x)
return 0;
if(p.y < a.s.y && p.y < a.t.y)
return 0;
if(p.y > a.s.y && p.y > a.t.y)
return 0;
return 1;
}
int intersection(Seg a, Seg b) {
if(onLine(a, b.s) || onLine(a, b.t) ||
onLine(b, a.s) || onLine(b, a.t))
return 1;
if(cross(a.s, a.t, b.s)*cross(a.s, a.t, b.t) < 0 &&
cross(a.t, a.s, b.s)*cross(a.t, a.s, b.t) < 0 &&
cross(b.s, b.t, a.s)*cross(b.s, b.t, a.t) < 0 &&
cross(b.t, b.s, a.s)*cross(b.t, b.s, a.t) < 0)
return 1;
return 0;
}
int g[2005][4005], gt[2005];
void addEdge(int x, int y) {
//printf("%d -> %d\n", x, y);
//g[x].push_back(y);
g[x][gt[x]++] = y;
}
int vfind[2005], findIdx;
int stk[2005], stkIdx;
int in_stk[2005], visited[2005];
int vgroup[2005];
int scc_cnt;
int scc(int nd) {
//printf("%d\n", nd);
in_stk[nd] = visited[nd] = 1;
stk[++stkIdx] = nd;
vfind[nd] = ++findIdx;
int mn = vfind[nd];
/*for(vector<int>::iterator it = g[nd].begin();
it != g[nd].end(); it++) {
if(!visited[*it])
mn = min(mn, scc(*it));
if(in_stk[*it])
mn = min(mn, vfind[*it]);
}*/
for(int i = 0; i < gt[nd]; i++) {
if(!visited[g[nd][i]]) {
mn = min(mn, scc(g[nd][i]));
//printf("%d --> %d\n", nd, g[nd][i]);
}
if(in_stk[g[nd][i]]){
mn = min(mn, vfind[g[nd][i]]);
// printf("%d %d %d\n", nd, g[nd][i], vfind[g[nd][i]]);
}
}
//printf("%d : %d %d\n", nd, mn, vfind[nd]);
if(mn == vfind[nd]) {
do {
in_stk[stk[stkIdx]] = 0;
vgroup[stk[stkIdx]] = scc_cnt;
//printf("%d ", stk[stkIdx]);
} while(stk[stkIdx--] != nd);
scc_cnt++;
}
return mn;
}
int main() {
int n;
int i, j;
Pt p[4], rect[1000][10];
while(scanf("%d", &n) == 1 && n) {
for(i = 0; i < n; i++) {
for(j = 0; j < 4; j++)
scanf("%f %f", &p[j].x, &p[j].y);
convexHull(4, p, rect[i]);
}
memset(visited, 0, sizeof(visited));
memset(in_stk, 0, sizeof(in_stk));
memset(gt, 0, sizeof(gt));
//for(i = 2*n; i >= 0; i--)
// g[i].clear();
for(i = 0; i < n; i++) {
//printf("%lf %lf %lf %lf\n", rect[i][0].x, rect[i][0].y, rect[i][2].x, rect[i][2].y);
//printf("%lf %lf %lf %lf\n\n", rect[i][1].x, rect[i][1].y, rect[i][3].x, rect[i][3].y);
//diagonal val(i<<1), another diagonal !val(i<<1|1)
for(j = 0; j < i; j++) {
Seg a, b;
a.s = rect[i][0], a.t = rect[i][2];
b.s = rect[j][0], b.t = rect[j][2];
int f1, f2, f3, f4;
f1 = intersection(a, b);
a.s = rect[i][0], a.t = rect[i][2];
b.s = rect[j][1], b.t = rect[j][3];
f2 = intersection(a, b);
a.s = rect[i][1], a.t = rect[i][3];
b.s = rect[j][0], b.t = rect[j][2];
f3 = intersection(a, b);
a.s = rect[i][1], a.t = rect[i][3];
b.s = rect[j][1], b.t = rect[j][3];
f4 = intersection(a, b);
if(f1) {
addEdge(i<<1, j<<1|1);
addEdge(j<<1, i<<1|1);
}
if(f2) {
addEdge(i<<1, j<<1);
addEdge(j<<1|1, i<<1|1);
}
if(f3) {
addEdge(i<<1|1, j<<1|1);
addEdge(j<<1, i<<1);
}
if(f4) {
addEdge(i<<1|1, j<<1);
addEdge(j<<1|1, i<<1);
}
}
}
scc_cnt = 1;
for(i = 0; i < 2*n; i++) {
if(!visited[i]) {
stkIdx = -1;
findIdx = 0;
scc(i);
}
}
int hasSolution = 1;
for(i = 0; i < n; i++)
if(vgroup[i<<1] == vgroup[i<<1|1])
hasSolution = 0;
puts(hasSolution ? "YES" : "NO");
}
return 0;
}
/*
2-SAT
ex. (A | B)&(!A | B) has a solution?
*/