SRM 587 DIV1

550

结论：同一层的交点共线。

很容易猜到，也可以跑几组数据验证。

利用结论就可以按层算，再利用对称性简化计算。

 1 using namespace std;

 2 #define maxn 70100

 3 class TriangleXor {

 4 public:

 5     int theArea(int);

 6 };

 7 double l[maxn] , x[maxn] , y[maxn] , H;

 8 int idx;

 9 

10 void addpoint(double k,double w) {// y = kx + 1 , y = x/w

11     double X = w/(1.0-k*w);

12     double Y = X/w;

13     x[++idx] = X;

14     y[idx] =Y;

15 }

16 

17 double dist(double x0,double y0,double x1,double y1) {

18     return sqrt((x0-x1)*(x0-x1)+(y0-y1)*(y0-y1));

19 }

20 

21 int TriangleXor::theArea(int w) {

22     double ans = 0.0;

23     if (w%2==0) {

24         ans = 0.5 * w / 2.0;

25     }

26     x[0] = y[0] = 0;

27     for (int i=1 ; i<=w ; i++ ) {

28         addpoint(-1.0/(double)i,w);

29     }

30     // H = (double)w / sqrt(1.0+w*w);

31     for (int i=0 ; i<w ; i+=2 ) {

32         if (i+1<=w) {

33             // double tmp = dist(x[i],y[i],x[i+1],y[i+1]);

34             // ans += tmp*H;

35             H = x[i+1]-x[i];

36             ans += H;

37         }

38     }

39     for (int i=0 ; i<w ; i+=2 ) if (i+2<=w) {

40         H = y[i+2] - y[i];

41         double len = 2.0 * (0.5-y[i+1]) * (double)w;

42 //        double len = w-2*x[i+1];

43         ans += len * H / 2.0;

44     }

45     cout<<ans<<endl;

46     return (int)floor(ans);

47 }

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 900

结论：用01串表示"NZ",任意两行（两列）要么相等，要么取反。

利用结论，贪心地在每一位枚举并检验，枚举复杂度O(n^2) ,检验复杂度O(n^3),总复杂度O(n^5)。

检验方法： g[i][j]表示第i,j行是否相同,利用cells求出部分i,用类似传递闭包的办法递推所有能求出来的关系，同时检验是否会有矛盾。

 1 class ThreeColorability {

 2 public:

 3     vector <string> lexSmallest(vector <string>);

 4 };

 5 #define maxn 55

 6 int g[maxn][maxn],n,m;

 7 

 8 int check(vector<string> c) {

 9     memset(g,-1,sizeof(g));

10     for (int i=0; i<n; i++ ) {

11         g[i][i] = 0;

12         for (int j=i+1; j<n; j++ ) {

13             for (int k=0; k<m; k++ ) if (c[i][k]!='?' && c[j][k]!='?') {

14                 if (g[i][j]==-1) {

15                     g[i][j] = (c[i][k]!=c[j][k]);

16                     g[j][i] = g[i][j];

17                 } else {

18                     if (g[i][j] && c[i][k]==c[j][k]) return 0;

19                     if (!g[i][j] && c[i][k]!=c[j][k]) return 0;

20                 }

21             }

22         }

23     }

24     for (int i=0; i<n; i++ )

25     for (int j=0; j<n; j++ )

26     for (int k=0; k<n; k++ )

27     if (g[j][i]!=-1 && g[i][k]!=-1) {

28         if (g[j][k]==-1) g[j][k] = g[j][i]^g[i][k];

29         else if (g[j][k] != (g[j][i]^g[i][k])) return 0;

30     }

31     return 1;

32 }

33 

34 vector <string> ThreeColorability::lexSmallest(vector <string> c) {

35     n = c.size();

36     m = c[0].size();

37     memset(g,-1,sizeof(g));

38     

39     if (!check(c)) return {};

40     

41     for (int i=0; i<n; i++ ) for (int j=0; j<m; j++ ) if (c[i][j]=='?') {

42         c[i][j]='N';

43         if (!check(c)) c[i][j] = 'Z';

44     }

45     return c;

46 }

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