zoj 2193 poj 2585 Window Pains

拓扑排序。

深刻体会：ACM比赛的精髓之处不在于学了某个算法或数据结构，而在于知道这个知识点但不知道这个问题可以用这个知识去解决！一看题目，根本想不到是拓扑排序。T_T......

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<algorithm>
using namespace std;
int mapp[50][50]; char s[1000];
int i, j, k, f;
vector<int> abc[1000];
int jz[20][20];
int c[20];
int ff[100][100];
void chushihua()
{
    jz[1][1] = 1;
    jz[2][1] = 1; jz[2][2] = 1;
    jz[3][2] = 1; jz[3][3] = 1;
    jz[4][3] = 1;
    jz[5][1] = 1; jz[5][4] = 1;
    jz[6][1] = 1; jz[6][2] = 1; jz[6][4] = 1; jz[6][5] = 1;
    jz[7][2] = 1; jz[7][3] = 1; jz[7][5] = 1; jz[7][6] = 1;
    jz[8][3] = 1; jz[8][6] = 1;
    jz[9][4] = 1; jz[9][7] = 1;
    jz[10][4] = 1; jz[10][5] = 1; jz[10][7] = 1; jz[10][8] = 1;
    jz[11][5] = 1; jz[11][6] = 1; jz[11][8] = 1; jz[11][9] = 1;
    jz[12][6] = 1; jz[12][9] = 1;
    jz[13][7] = 1;
    jz[14][7] = 1; jz[14][8] = 1;
    jz[15][8] = 1; jz[15][9] = 1;
    jz[16][9] = 1;
}
int main()
{
    memset(jz, 0, sizeof(jz));
    chushihua();int ge;
    while (~scanf("%s", s))
    {
        for (i = 0; i < 20; i++)abc[i].clear();
        memset(ff, 0, sizeof(ff));
        if (strcmp("ENDOFINPUT", s) == 0) break;
        memset(c, 0, sizeof(c));
        for (i = 1; i <= 4; i++) for (j = 1; j <= 4; j++) scanf("%d", &mapp[i][j]);
        for (i = 1; i <= 4; i++)
        {
            for (j = 1; j <= 4; j++)
            {
                ge = (i - 1) * 4 + j;
                for (f = 1; f <= 9; f++)
                {
                    if (jz[ge][f] == 1 && f != mapp[i][j])
                    {
                        if (ff[mapp[i][j]][f] == 0)
                        {
                            ff[mapp[i][j]][f] = 1;
                            abc[mapp[i][j]].push_back(f);
                            c[f]++;
                        }    
                    }
                }
            }
        }
        scanf("%s", s); int df, summ = 0;
        while (1)
        {
            df = 0;
            for (i = 1; i <= 9; i++)
            {
                if (c[i] == 0)
                {
                    c[i]--;df = 1;summ++;
                    for (j = 0; j < abc[i].size(); j++) c[abc[i][j]]--;
                    break;
                }
            }
            if (summ == 9){printf("THESE WINDOWS ARE CLEAN
");break;}
            if (df == 0){printf("THESE WINDOWS ARE BROKEN
");break;}
        }
    }
    return 0;
}