hdu 5749 Colmerauer
题意:对于给定的$n \times m$矩阵$M$,定义$S(a,b)$为$M$的所有$a \times b$子矩阵的权重之和。一个矩阵的权重是指矩阵中所有马鞍点权值之和,在一个矩阵中某点是马鞍点当且仅当它在所在行是唯一一个最小的,同时在所在列中是唯一一个最大的。现在输入矩阵$M$,要求计算$W= \sum\sum{abS(a,b)}, 1 \leq a \leq n, 1 \leq b \leq m$。数据范围$1 \leq n, m \leq 1000, 0 \leq M(i, j) \leq 1000000$。
分析: 考虑每个马鞍点对答案$W$的贡献,答案可以重写为每个马鞍点的权值与其所在子矩阵面积的和的积的和。考虑位于$i$行$j$列的点,我们设在$j$列从该点出发向上连续的$y_1$个点中除了点$(i, j)$其余点对应的权值均严格小于该点权值,并假设$y_1$是最大的。记$y1$为该点的最大向上关联长度,同理我们可以得到向下,向左,向右的最大关联长度,显然这些长度至少为$1$,按照左右上下的顺序得到关于$(i, j)$的四元组$(x_1, x_2, y_1, y_2)$,把四个方向对应到坐标轴方向,该点对应到原点,可以得到四个所谓的象限。那么如果该点在某个子矩阵内为马鞍点,必然有该子矩阵左上顶点在第二象限,右下顶点在第四象限(可以通过画图很直观得看出来)。那么所有子矩阵的面积之和为$g(i, j) = \sum\sum\sum\sum{(p + q + 1)(r + s + 1)}$其中$p, q, r, s$分别独立地在$[0, x_1), [0, x_2), [0, y_1), [0, y_2)$中取值,那么不难将其展开得到关于$x_1, x_2, y_1, y_2$的代数表达式。那么最终答案应该是$\sum\sum{g(i, j)w(i, j)}$,$w$是权值函数。于是现在只需要对每个点计算其四元组,可以采用rmq+二分的方法(尽管这种做法并非最优)。这样我们算法的复杂度就是$O(n^2log(n))$。取模使用unsigned int自然溢出就可以了。代码如下:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <algorithm> 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <queue> 6 #include <map> 7 #include <set> 8 #include <stack> 9 #include <ctime> 10 #include <cmath> 11 #include <iostream> 12 #include <assert.h> 13 #define PI acos(-1.) 14 #pragma comment(linker, "/STACK:102400000,102400000") 15 #define max(a, b) ((a) > (b) ? (a) : (b)) 16 #define min(a, b) ((a) < (b) ? (a) : (b)) 17 #define mp std :: make_pair 18 #define st first 19 #define nd second 20 #define keyn (root->ch[1]->ch[0]) 21 #define lson (u << 1) 22 #define rson (u << 1 | 1) 23 #define pii std :: pair<int, int> 24 #define pll pair<ll, ll> 25 #define pb push_back 26 #define type(x) __typeof(x.begin()) 27 #define foreach(i, j) for(type(j)i = j.begin(); i != j.end(); i++) 28 #define FOR(i, s, t) for(int i = (s); i <= (t); i++) 29 #define ROF(i, t, s) for(int i = (t); i >= (s); i--) 30 #define dbg(x) std::cout << x << std::endl 31 #define dbg2(x, y) std::cout << x << " " << y << std::endl 32 #define clr(x, i) memset(x, (i), sizeof(x)) 33 #define maximize(x, y) x = max((x), (y)) 34 #define minimize(x, y) x = min((x), (y)) 35 using namespace std; 36 typedef long long ll; 37 const int int_inf = 0x3f3f3f3f; 38 const ll ll_inf = 0x3f3f3f3f3f3f3f3f; 39 const int INT_INF = (int)((1ll << 31) - 1); 40 const double double_inf = 1e30; 41 const double eps = 1e-14; 42 typedef unsigned long long ul; 43 typedef unsigned int ui; 44 inline int readint(){ 45 int x; 46 scanf("%d", &x); 47 return x; 48 } 49 inline int readstr(char *s){ 50 scanf("%s", s); 51 return strlen(s); 52 } 53 //Here goes 2d geometry templates 54 struct Point{ 55 double x, y; 56 Point(double x = 0, double y = 0) : x(x), y(y) {} 57 }; 58 typedef Point Vector; 59 Vector operator + (Vector A, Vector B){ 60 return Vector(A.x + B.x, A.y + B.y); 61 } 62 Vector operator - (Point A, Point B){ 63 return Vector(A.x - B.x, A.y - B.y); 64 } 65 Vector operator * (Vector A, double p){ 66 return Vector(A.x * p, A.y * p); 67 } 68 Vector operator / (Vector A, double p){ 69 return Vector(A.x / p, A.y / p); 70 } 71 bool operator < (const Point& a, const Point& b){ 72 return a.x < b.x || (a.x == b.x && a.y < b.y); 73 } 74 int dcmp(double x){ 75 if(abs(x) < eps) return 0; 76 return x < 0 ? -1 : 1; 77 } 78 bool operator == (const Point& a, const Point& b){ 79 return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0; 80 } 81 double Dot(Vector A, Vector B){ 82 return A.x * B.x + A.y * B.y; 83 } 84 double Len(Vector A){ 85 return sqrt(Dot(A, A)); 86 } 87 double Angle(Vector A, Vector B){ 88 return acos(Dot(A, B) / Len(A) / Len(B)); 89 } 90 double Cross(Vector A, Vector B){ 91 return A.x * B.y - A.y * B.x; 92 } 93 double Area2(Point A, Point B, Point C){ 94 return Cross(B - A, C - A); 95 } 96 Vector Rotate(Vector A, double rad){ 97 //rotate counterclockwise 98 return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad)); 99 } 100 Vector Normal(Vector A){ 101 double L = Len(A); 102 return Vector(-A.y / L, A.x / L); 103 } 104 void Normallize(Vector &A){ 105 double L = Len(A); 106 A.x /= L, A.y /= L; 107 } 108 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w){ 109 Vector u = P - Q; 110 double t = Cross(w, u) / Cross(v, w); 111 return P + v * t; 112 } 113 double DistanceToLine(Point P, Point A, Point B){ 114 Vector v1 = B - A, v2 = P - A; 115 return abs(Cross(v1, v2)) / Len(v1); 116 } 117 double DistanceToSegment(Point P, Point A, Point B){ 118 if(A == B) return Len(P - A); 119 Vector v1 = B - A, v2 = P - A, v3 = P - B; 120 if(dcmp(Dot(v1, v2)) < 0) return Len(v2); 121 else if(dcmp(Dot(v1, v3)) > 0) return Len(v3); 122 else return abs(Cross(v1, v2)) / Len(v1); 123 } 124 Point GetLineProjection(Point P, Point A, Point B){ 125 Vector v = B - A; 126 return A + v * (Dot(v, P - A) / Dot(v, v)); 127 } 128 bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){ 129 //Line1:(a1, a2) Line2:(b1,b2) 130 double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1), 131 c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1); 132 return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0; 133 } 134 bool OnSegment(Point p, Point a1, Point a2){ 135 return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 -p)) < 0; 136 } 137 Vector GetBisector(Vector v, Vector w){ 138 Normallize(v), Normallize(w); 139 return Vector((v.x + w.x) / 2, (v.y + w.y) / 2); 140 } 141 142 bool OnLine(Point p, Point a1, Point a2){ 143 Vector v1 = p - a1, v2 = a2 - a1; 144 double tem = Cross(v1, v2); 145 return dcmp(tem) == 0; 146 } 147 struct Line{ 148 Point p; 149 Vector v; 150 Point point(double t){ 151 return Point(p.x + t * v.x, p.y + t * v.y); 152 } 153 Line(Point p, Vector v) : p(p), v(v) {} 154 }; 155 struct Circle{ 156 Point c; 157 double r; 158 Circle(Point c, double r) : c(c), r(r) {} 159 Circle(int x, int y, int _r){ 160 c = Point(x, y); 161 r = _r; 162 } 163 Point point(double a){ 164 return Point(c.x + cos(a) * r, c.y + sin(a) * r); 165 } 166 }; 167 int GetLineCircleIntersection(Line L, Circle C, double &t1, double& t2, std :: vector<Point>& sol){ 168 double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y; 169 double e = a * a + c * c, f = 2 * (a * b + c * d), g = b * b + d * d - C.r * C.r; 170 double delta = f * f - 4 * e * g; 171 if(dcmp(delta) < 0) return 0; 172 if(dcmp(delta) == 0){ 173 t1 = t2 = -f / (2 * e); sol.pb(L.point(t1)); 174 return 1; 175 } 176 t1 = (-f - sqrt(delta)) / (2 * e); sol.pb(L.point(t1)); 177 t2 = (-f + sqrt(delta)) / (2 * e); sol.pb(L.point(t2)); 178 return 2; 179 } 180 double angle(Vector v){ 181 return atan2(v.y, v.x); 182 //(-pi, pi] 183 } 184 int GetCircleCircleIntersection(Circle C1, Circle C2, std :: vector<Point>& sol){ 185 double d = Len(C1.c - C2.c); 186 if(dcmp(d) == 0){ 187 if(dcmp(C1.r - C2.r) == 0) return -1; //two circle duplicates 188 return 0; //two circles share identical center 189 } 190 if(dcmp(C1.r + C2.r - d) < 0) return 0; //too close 191 if(dcmp(abs(C1.r - C2.r) - d) > 0) return 0; //too far away 192 double a = angle(C2.c - C1.c); // angle of vector(C1, C2) 193 double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d)); 194 Point p1 = C1.point(a - da), p2 = C1.point(a + da); 195 sol.pb(p1); 196 if(p1 == p2) return 1; 197 sol.pb(p2); 198 return 2; 199 } 200 int GetPointCircleTangents(Point p, Circle C, Vector* v){ 201 Vector u = C.c - p; 202 double dist = Len(u); 203 if(dist < C.r) return 0;//p is inside the circle, no tangents 204 else if(dcmp(dist - C.r) == 0){ 205 // p is on the circles, one tangent only 206 v[0] = Rotate(u, PI / 2); 207 return 1; 208 }else{ 209 double ang = asin(C.r / dist); 210 v[0] = Rotate(u, -ang); 211 v[1] = Rotate(u, +ang); 212 return 2; 213 } 214 } 215 int GetCircleCircleTangents(Circle A, Circle B, Point* a, Point* b){ 216 //a[i] store point of tangency on Circle A of tangent i 217 //b[i] store point of tangency on Circle B of tangent i 218 //six conditions is in consideration 219 int cnt = 0; 220 if(A.r < B.r) { std :: swap(A, B); std :: swap(a, b); } 221 int d2 = (A.c.x - B.c.x) * (A.c.x - B.c.x) + (A.c.y - B.c.y) * (A.c.y - B.c.y); 222 int rdiff = A.r - B.r; 223 int rsum = A.r + B.r; 224 if(d2 < rdiff * rdiff) return 0; // one circle is inside the other 225 double base = atan2(B.c.y - A.c.y, B.c.x - A.c.x); 226 if(d2 == 0 && A.r == B.r) return -1; // two circle duplicates 227 if(d2 == rdiff * rdiff){ // internal tangency 228 a[cnt] = A.point(base); b[cnt] = B.point(base); cnt++; 229 return 1; 230 } 231 double ang = acos((A.r - B.r) / sqrt(d2)); 232 a[cnt] = A.point(base + ang); b[cnt++] = B.point(base + ang); 233 a[cnt] = A.point(base - ang); b[cnt++] = B.point(base - ang); 234 if(d2 == rsum * rsum){ 235 //one internal tangent 236 a[cnt] = A.point(base); 237 b[cnt++] = B.point(base + PI); 238 }else if(d2 > rsum * rsum){ 239 //two internal tangents 240 double ang = acos((A.r + B.r) / sqrt(d2)); 241 a[cnt] = A.point(base + ang); b[cnt++] = B.point(base + ang + PI); 242 a[cnt] = A.point(base - ang); b[cnt++] = B.point(base - ang + PI); 243 } 244 return cnt; 245 } 246 Point ReadPoint(){ 247 double x, y; 248 scanf("%lf%lf", &x, &y); 249 return Point(x, y); 250 } 251 Circle ReadCircle(){ 252 double x, y, r; 253 scanf("%lf%lf%lf", &x, &y, &r); 254 return Circle(x, y, r); 255 } 256 //Here goes 3d geometry templates 257 struct Point3{ 258 double x, y, z; 259 Point3(double x = 0, double y = 0, double z = 0) : x(x), y(y), z(z) {} 260 }; 261 typedef Point3 Vector3; 262 Vector3 operator + (Vector3 A, Vector3 B){ 263 return Vector3(A.x + B.x, A.y + B.y, A.z + B.z); 264 } 265 Vector3 operator - (Vector3 A, Vector3 B){ 266 return Vector3(A.x - B.x, A.y - B.y, A.z - B.z); 267 } 268 Vector3 operator * (Vector3 A, double p){ 269 return Vector3(A.x * p, A.y * p, A.z * p); 270 } 271 Vector3 operator / (Vector3 A, double p){ 272 return Vector3(A.x / p, A.y / p, A.z / p); 273 } 274 double Dot3(Vector3 A, Vector3 B){ 275 return A.x * B.x + A.y * B.y + A.z * B.z; 276 } 277 double Len3(Vector3 A){ 278 return sqrt(Dot3(A, A)); 279 } 280 double Angle3(Vector3 A, Vector3 B){ 281 return acos(Dot3(A, B) / Len3(A) / Len3(B)); 282 } 283 double DistanceToPlane(const Point3& p, const Point3 &p0, const Vector3& n){ 284 return abs(Dot3(p - p0, n)); 285 } 286 Point3 GetPlaneProjection(const Point3 &p, const Point3 &p0, const Vector3 &n){ 287 return p - n * Dot3(p - p0, n); 288 } 289 Point3 GetLinePlaneIntersection(Point3 p1, Point3 p2, Point3 p0, Vector3 n){ 290 Vector3 v = p2 - p1; 291 double t = (Dot3(n, p0 - p1) / Dot3(n, p2 - p1)); 292 return p1 + v * t;//if t in range [0, 1], intersection on segment 293 } 294 Vector3 Cross(Vector3 A, Vector3 B){ 295 return Vector3(A.y * B.z - A.z * B.y, A.z * B.x - A.x * B.z, A.x * B.y - A.y * B.x); 296 } 297 double Area3(Point3 A, Point3 B, Point3 C){ 298 return Len3(Cross(B - A, C - A)); 299 } 300 class cmpt{ 301 public: 302 bool operator () (const int &x, const int &y) const{ 303 return x > y; 304 } 305 }; 306 307 int Rand(int x, int o){ 308 //if o set, return [1, x], else return [0, x - 1] 309 if(!x) return 0; 310 int tem = (int)((double)rand() / RAND_MAX * x) % x; 311 return o ? tem + 1 : tem; 312 } 313 void data_gen(){ 314 srand(time(0)); 315 freopen("in.txt", "w", stdout); 316 int kases = 10; 317 printf("%d\n", kases); 318 while(kases--){ 319 int sz = 2e4; 320 int m = 1e5; 321 printf("%d %d\n", sz, m); 322 FOR(i, 1, sz) printf("%d ", Rand(100, 1)); 323 printf("\n"); 324 FOR(i, 1, sz) printf("%d ", Rand(1e9, 1)); 325 printf("\n"); 326 FOR(i, 1, m){ 327 int l = Rand(sz, 1); 328 int r = Rand(sz, 1); 329 int c = Rand(1e9, 1); 330 printf("%d %d %d %d\n", l, r, c, Rand(100, 1)); 331 } 332 } 333 } 334 335 struct cmpx{ 336 bool operator () (int x, int y) { return x > y; } 337 }; 338 const int maxn = 1e3 + 10; 339 ui mt[maxn][maxn]; 340 ui x1[maxn][maxn], x2[maxn][maxn], y1[maxn][maxn], y2[maxn][maxn]; 341 int n, m; 342 ui bg[maxn][15]; 343 ui query(int l, int r, int o){ 344 int len = (r - l + 1); 345 int i = 0; 346 while((1 << i) <= len) ++i; 347 --i; 348 int sp = r - (1 << i) + 1; 349 if(o) return max(bg[l][i], bg[sp][i]); 350 else return min(bg[l][i], bg[sp][i]); 351 } 352 int main(){ 353 //data_gen(); return 0; 354 //C(); return 0; 355 int debug = 0; 356 if(debug) freopen("in.txt", "r", stdin); 357 //freopen("out.txt", "w", stdout); 358 int T = readint(); 359 while(T--){ 360 scanf("%d%d", &n, &m); 361 FOR(i, 1, n) FOR(j, 1, m) scanf("%u", &mt[i][j]); 362 FOR(i, 1, n) FOR(j, 1, m) x1[i][j] = x2[i][j] = y1[i][j] = y2[i][j] = 1; 363 FOR(i, 1, n){ 364 FOR(j, 1, m) bg[j][0] = mt[i][j]; 365 for(int i = 1; (1 << i) <= m; i++){ 366 int len = 1 << i; 367 for(int j = 1; j + len - 1 <= m; j++) bg[j][i] = min(bg[j][i - 1], bg[j + len / 2][i - 1]); 368 } 369 FOR(j, 1, m){ 370 if(j == 1 || mt[i][j] >= mt[i][j - 1]) continue; 371 int l = 0, r = j - 1; 372 while(r - l > 1){ 373 int mid = (l + r) >> 1; 374 ui tem = query(mid, r, 0); 375 if(tem > mt[i][j]) r = mid; 376 else l = mid; 377 } 378 x1[i][j] = j - r + 1; 379 } 380 FOR(j, 1, m){ 381 if(j == m || mt[i][j] >= mt[i][j + 1]) continue; 382 int l = j + 1, r = m + 1; 383 while(r - l > 1){ 384 int mid = (l + r) >> 1; 385 ui tem = query(l, mid, 0); 386 if(tem > mt[i][j]) l = mid; 387 else r = mid; 388 } 389 x2[i][j] = l - j + 1; 390 } 391 } 392 FOR(j, 1, m){ 393 FOR(i, 1, n) bg[i][0] = mt[i][j]; 394 for(int i = 1; (1 << i) <= n; i++){ 395 int len = 1 << i; 396 for(int j = 1; j + len - 1 <= n; j++) bg[j][i] = max(bg[j][i - 1], bg[j + len / 2][i - 1]); 397 } 398 FOR(i, 1, n){ 399 if(i == 1 || mt[i][j] <= mt[i - 1][j]) continue; 400 int l = 0, r = i - 1; 401 while(r - l > 1){ 402 int mid = (l + r) >> 1; 403 ui tem = query(mid, r, 1); 404 if(tem < mt[i][j]) r = mid; 405 else l = mid; 406 } 407 y1[i][j] = i - r + 1; 408 } 409 FOR(i, 1, n){ 410 if(i == n || mt[i][j] <= mt[i + 1][j]) continue; 411 int l = i + 1, r = n + 1; 412 while(r - l > 1){ 413 int mid = (l + r) >> 1; 414 ui tem = query(l, mid, 1); 415 if(tem < mt[i][j]) l = mid; 416 else r = mid; 417 } 418 y2[i][j] = l - i + 1; 419 } 420 } 421 ui ans = 0; 422 FOR(i, 1, n) FOR(j, 1, m){ 423 ui X1 = x1[i][j] * (x1[i][j] - 1) / 2; 424 ui X2 = x2[i][j] * (x2[i][j] - 1) / 2; 425 ui Y1 = y1[i][j] * (y1[i][j] - 1) / 2; 426 ui Y2 = y2[i][j] * (y2[i][j] - 1) / 2; 427 ui _x1 = x1[i][j], _x2 = x2[i][j], _y1 = y1[i][j], _y2 = y2[i][j]; 428 ui para = _x2 * _y2 * X1 * Y1 + X1 * Y2 * _x2 * _y1 + X2 * Y1 * _x1 * _y2 + X2 * Y2 * _x1 * _y1; 429 para += X1 * _x2 * _y1 * _y2 + X2 * _x1 * _y1 * _y2 + Y1 * _x1 * _x2 * _y2 + Y2 * _x1 * _x2 * _y1; 430 para += _x1 * _y1 * _y2 * _x2; 431 ans += para * mt[i][j]; 432 } 433 printf("%u\n", ans); 434 } 435 return 0; 436 }