hdu 3231 Box Relations(拓撲排序)

Box Relations

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1108    Accepted Submission(s): 420
Special Judge

Problem Description

There are n boxes C₁, C₂, ..., C_n in 3D space. The edges of the boxes are parallel to the x, y or z-axis. We provide some relations of the boxes, and your task is to construct a set of boxes satisfying all these relations.

There are four kinds of relations (1 <= i,j <= n, i is different from j):

• I i j: The intersection volume of C_i and C_j is positive.
  
• X i j: The intersection volume is zero, and any point inside C_i has smaller x-coordinate than any point inside C_j.
  
• Y i j: The intersection volume is zero, and any point inside C_i has smaller y-coordinate than any point inside C_j.
  
• Z i j: The intersection volume is zero, and any point inside C_i has smaller z-coordinate than any point inside C_j.

.

 

Input

There will be at most 30 test cases. Each case begins with a line containing two integers n (1 <= n <= 1,000) and R (0 <= R <= 100,000), the number of boxes and the number of relations. Each of the following R lines describes a relation, written in the format above. The last test case is followed by n=R=0, which should not be processed.

 

Output

For each test case, print the case number and either the word POSSIBLE or IMPOSSIBLE. If it's possible to construct the set of boxes, the i-th line of the followingn lines contains six integers x1, y1, z1, x2, y2, z2, that means the i-th box is the set of points (x,y,z) satisfying x1 <= x <= x2, y1 <= y <= y2, z1 <= z <= z2. The absolute values of x1, y1, z1, x2, y2, z2 should not exceed 1,000,000.

Print a blank line after the output of each test case.

 

Sample Input

3 2 I 1 2 X 2 3 3 3 Z 1 2 Z 2 3 Z 3 1 1 0 0 0

 

Sample Output

Case 1: POSSIBLE 0 0 0 2 2 2 1 1 1 3 3 3 8 8 8 9 9 9 Case 2: IMPOSSIBLE Case 3: POSSIBLE 0 0 0 1 1 1

 

Source

2009 Asia Wuhan Regional Contest Hosted by Wuhan University

 

題意：三維座標下讓你構造滿足條件的n個長方體。有R個要求，I i j表示第i個長方體和第j個長方體必須有交集;   X  i  j 表示 i 任意一點的x座標必須小於 j 的;

            Y i j ,Z i j同理。

題解 拓撲排序。分別把三個面拆成三條線段，再拆成兩個點，根據要求構造關係，再進行三次拓撲排序即可。

#include<cstring>
#include<algorithm>
#include<cstdio>
#include<iostream>
#include<queue>
#include<set>
using namespace std;

const int N=2005;
const int M=2e5+5;

int n,m;
int c[N];
int topo[N],t;
vector<int>X[N],Y[N],Z[N];
int ans[3][N];

void init() {
    for(int i=0; i<=2*n; i++) {
        X[i].clear();
        Y[i].clear();
        Z[i].clear();
    }
    for(int i=0; i<n; i++) {
        X[i].push_back(i+n);
        Y[i].push_back(i+n);
        Z[i].push_back(i+n);
    }
}

bool dfs(int u,vector<int>G[]) {
    c[u]=-1;
    for(int v=0; v<G[u].size(); v++) {
        int to=G[u][v];
        if(c[to]<0)return false;
        else if(!c[to]&&!dfs(to,G))return false;
    }
    c[u]=1;
    topo[--t]=u;
    return true;
}

bool toposort(vector<int>G[]) {
    t=2*n;
    memset(c,0,sizeof (c));
    for(int u=0; u<2*n; u++) {
        if(!c[u])if(!dfs(u,G))return false;
    }
    return true;
}

bool solve() {
    if(!toposort(X))return false;
    for(int i=0; i<2*n; i++)
        ans[0][topo[i]]=i;
    if(!toposort(Y))return false;
    for(int i=0; i<2*n; i++)
        ans[1][topo[i]]=i;
    if(!toposort(Z))return false;
    for(int i=0; i<2*n; i++)
        ans[2][topo[i]]=i;
    return true;
}

int main() {
   // freopen("test.in","r",stdin);
    int ca=1;
    while(~scanf("%d%d",&n,&m)&&(n+m)) {
        char c[2];
        int x,y;
        init();
        while(m--) {
            scanf("%s%d%d",c,&x,&y);
            x--,y--;
            if(c[0]=='I') {
                X[x].push_back(y+n);
                X[y].push_back(x+n);
                Y[x].push_back(y+n);
                Y[y].push_back(x+n);
                Z[x].push_back(y+n);
                Z[y].push_back(x+n);
            } else if(c[0]=='X') {
                X[x+n].push_back(y);
            } else if(c[0]=='Y') {
                Y[n+x].push_back(y);
            } else {
                Z[n+x].push_back(y);
            }
        }
        if(!solve()) {
            printf("Case %d: IMPOSSIBLE\n\n",ca++);
            continue;
        }
        printf("Case %d: POSSIBLE\n",ca++);
        for(int i=0; i<n; i++) {
            printf("%d %d %d %d %d %d\n",ans[0][i],ans[1][i],ans[2][i],ans[0][i+n],ans[1][i+n],ans[2][i+n]);
        }
        putchar(10);
    }
    return 0;
}