拓扑排序(好题)hdu3231

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Box Relations

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 974    Accepted Submission(s): 356
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 2I 1 2X 2 33 3Z 1 2Z 2 3Z 3 11 00 0

 

Sample Output

Case 1: POSSIBLE0 0 0 2 2 21 1 1 3 3 38 8 8 9 9 9Case 2: IMPOSSIBLECase 3: POSSIBLE0 0 0 1 1 1

思路：每个盒子有六个面，i表示上表面，i+n表示下表面，然后根据给的关系建图，I表示a的上表面在b的下表面上面，a的下表面在b的上表面下面，x,y,z的操作相同，然后把x,y,z三维分别进行拓扑排序，如果能够进行得话，就说明存在这样的顺序。

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=2010;const int dimension=3;int n,R;vector<int> g[3][maxn];int in[dimension][maxn];void init(){    memset(in,0,sizeof(in));    for(int i=0;i<dimension;i++)        for(int j=0;j<=2*n;j++)g[i][j].clear();    for(int i=0;i<dimension;i++)        for(int j=1;j<=n;j++)        {            g[i][j].push_back(j+n);            in[i][j+n]++;        }}bool topo(int x[],int d){    queue<int> q;    for(int i=1;i<=2*n;i++)        if(!in[d][i])            q.push(i);    int cnt=0;    while(!q.empty())    {        int b=q.front();        q.pop();        x[b]=cnt++;        int len=g[d][b].size();        for(int i=0;i<len;i++)            if(--in[d][g[d][b][i]]==0)q.push(g[d][b][i]);    }    for(int i=1;i<=2*n;i++)        if(in[d][i])return false;    return true;}void solve(){    int ans[dimension][maxn];    for(int i=0;i<dimension;i++)        if(!topo(ans[i],i)){printf("IMPOSSIBLE\n");return;}    printf("POSSIBLE\n");    for(int i=1;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]);}int main(){    int cas=1;    while(scanf("%d%d",&n,&R)!=EOF,n||R)    {        init();        while(R--)        {            char op[5];            int x,y;            scanf("%s%d%d",op,&x,&y);            if(op[0]=='I')            {                for(int i=0;i<dimension;i++)                {                    g[i][x].push_back(y+n);                    in[i][y+n]++;                    g[i][y].push_back(x+n);                    in[i][x+n]++;                }            }            else            {                g[op[0]-'X'][x+n].push_back(y);                in[op[0]-'X'][y]++;            }        }        printf("Case %d: ",cas++);        solve();        printf("\n");    }    return 0;}