poj 2585 Window Pains （建圖+拓撲排序）

Window Pains Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1357   Accepted: 679 Description Boudreaux likes to multitask, especially when it comes to using his computer. Never satisfied with just running one application at a time, he usually runs nine applications, each in its own window. Due to limited screen real estate, he overlaps these windows and brings whatever window he currently needs to work with to the foreground. If his screen were a 4 x 4 grid of squares, each of Boudreaux's windows would be represented by the following 2 x 2 windows:  1 1 . . 1 1 . . . . . . . . . . . 2 2 . . 2 2 . . . . . . . . . . . 3 3 . . 3 3 . . . . . . . . . . . . 4 4 . . 4 4 . . . . . . . . . . . 5 5 . . 5 5 . . . . . . . . . . . 6 6 . . 6 6 . . . . . . . . . . . . 7 7 . . 7 7 . . . . . . . . . . . 8 8 . . 8 8 . . . . . . . . . . . 9 9 . . 9 9 When Boudreaux brings a window to the foreground, all of its squares come to the top, overlapping any squares it shares with other windows. For example, if window 1and then window 2 were brought to the foreground, the resulting representation would be: 1 2 2 ? 1 2 2 ? ? ? ? ? ? ? ? ? If window 4 were then brought to the foreground: 1 2 2 ? 4 4 2 ? 4 4 ? ? ? ? ? ? . . . and so on . . .  Unfortunately, Boudreaux's computer is very unreliable and crashes often. He could easily tell if a crash occurred by looking at the windows and seeing a graphical representation that should not occur if windows were being brought to the foreground correctly. And this is where you come in . . . Input Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets.  A single data set has 3 components:  Start line - A single line:  START  Screen Shot - Four lines that represent the current graphical representation of the windows on Boudreaux's screen. Each position in this 4 x 4 matrix will represent the current piece of window showing in each square. To make input easier, the list of numbers on each line will be delimited by a single space.  End line - A single line:  END  After the last data set, there will be a single line:  ENDOFINPUT  Note that each piece of visible window will appear only in screen areas where the window could appear when brought to the front. For instance, a 1 can only appear in the top left quadrant. Output For each data set, there will be exactly one line of output. If there exists a sequence of bringing windows to the foreground that would result in the graphical representation of the windows on Boudreaux's screen, the output will be a single line with the statement:  THESE WINDOWS ARE CLEAN  Otherwise, the output will be a single line with the statement:  THESE WINDOWS ARE BROKEN  Sample Input START1 2 3 34 5 6 67 8 9 97 8 9 9ENDSTART1 1 3 34 1 3 37 7 9 97 7 9 9ENDENDOFINPUT Sample Output THESE WINDOWS ARE CLEANTHESE WINDOWS ARE BROKEN Source South Central USA 2003

題意：給定視窗疊放後的狀態  讓你判斷電腦是否死機   

問題轉化：對1~9號視窗進行拓撲排序，如果合理則未死機，反之則反。 

建圖思路：對方塊進行分析，一個方塊被多個視窗覆蓋，方塊顯示的是最上面的視窗，則可以將其他的視窗與最上面的建立大小關係（建邊）。

代碼：

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <stack>#define maxn 15using namespace std;int n,m,ans;int a[10][10];int node[5][5][5];int num[maxn];char s[50];vector<int>v[maxn];stack<int>sta;void addedge()          // 建邊{    int i,j,k,nx;    memset(num,0,sizeof(num));    for(i=1; i<=4; i++)    {        for(j=1; j<=4; j++)        {            nx=a[i][j];            for(k=1;node[i][j][k]!=0;k++)            {                if(node[i][j][k]!=nx)                {                    num[node[i][j][k]]++;                    v[nx].push_back(node[i][j][k]);                }            }        }    }}bool tuopusort(){    int i,j,nx,cnt=0,sz;    while(!sta.empty()) sta.pop();    for(i=1;i<=9;i++)    {        if(num[i]==0) sta.push(i);    }    while(!sta.empty())    {        nx=sta.top();        cnt++;        sta.pop();        sz=v[nx].size();        for(i=0;i<sz;i++)        {            num[v[nx][i]]--;            if(num[v[nx][i]]==0) sta.push(v[nx][i]);        }    }    if(cnt==9) return true ;    return false ;}int main(){    int i,j;    memset(node,0,sizeof(node));    node[1][2][1]=1;    node[1][2][2]=2;    // 對每個方塊可能顯示的視窗打表    node[1][3][1]=2;    node[1][3][2]=3;    node[2][1][1]=1;    node[2][1][2]=4;    node[2][2][1]=1;    node[2][2][2]=2;    node[2][2][3]=4;    node[2][2][4]=5;    node[2][3][1]=2;    node[2][3][2]=3;    node[2][3][3]=5;    node[2][3][4]=6;    node[2][4][1]=3;    node[2][4][2]=6;    node[3][1][1]=4;    node[3][1][2]=7;    node[3][2][1]=4;    node[3][2][2]=5;    node[3][2][3]=7;    node[3][2][4]=8;    node[3][3][1]=5;    node[3][3][2]=6;    node[3][3][3]=8;    node[3][3][4]=9;    node[3][4][1]=6;    node[3][4][2]=9;    node[4][2][1]=7;    node[4][2][2]=8;    node[4][3][1]=8;    node[4][3][2]=9;    while(scanf("%s",s),strcmp(s,"ENDOFINPUT")!=0)    {        for(i=1; i<=4; i++)        {            for(j=1; j<=4; j++)            {                scanf("%d",&a[i][j]);            }        }        for(i=1; i<=9; i++)        {            v[i].clear();        }        addedge();        if(tuopusort()) printf("THESE WINDOWS ARE CLEAN\n");        else printf("THESE WINDOWS ARE BROKEN\n");        scanf("%s",s);    }    return 0;}