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 }
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 }
SRM 587 DIV1
时间： 2024-11-05 14:51:22