SRM 442

相比起443来略为简单。D1三道题分别是简单题、简单题和简单题。

 

D1 250p  Underprimes

 

没什么好说的，分解质因数，拼手速的题目。

 

D1 500p  BedroomFloor

 

模拟题，有些麻烦。

 

首先算出中间完整的正方形个数。然后用边上完整的正方形去和原长方形做交集运算，分出各种小矩形。

 

最后用4和1相拼，3和2相拼，3和1相拼，2和1相拼，1和1相拼。

 

class BedroomFloor { long long a[6]; void add ( int x, int y, int x1, int y1, int x2, int y2 ) { //printf("add: (%d,%d) in (%d,%d)×(%d,%d)/n", x, y, x1, y1, x2, y2); x1 = Max ( x, x1 ), y1 = Max ( y, y1 ), x2 = Min ( x + 5, x2 ), y2 = Min ( y + 5, y2 ); if ( x1 >= x2 || y1 >= y2 ) return; if ( ( ( x / 5 ) + ( y / 5 ) ) & 1 ) { //printf(" -> %d of length %d/n", x2 - x1, y2 - y1 ); a[y2 - y1] += x2 - x1; } else { a[x2 - x1] += y2 - y1; //printf(" -> %d of length %d/n", y2 - y1, x2 - x1 ); } } public: long long numberOfSticks(int x1, int y1, int x2, int y2) { memset ( a, 0, sizeof ( a ) ); // translation int t = x1 / 10 * 10; x1 -= t, x2 -= t; t = y1 / 10 * 10; y1 -= t, y2 -= t; int x, y, xx = x2 / 10 * 10, yy = y2 / 10 * 10; if ( x2 > 10 && y2 > 10 ) a[5] += ( long long ) ( xx - 10 ) * ( long long ) ( yy - 10 ) / 5; //printf ( "%lld/n", a[5] ); for ( x = 0; x < x2 + 10; x += 10 ) { //printf ( "1/n" ); add ( x, 0, x1, y1, x2, y2 ); add ( x, 5, x1, y1, x2, y2 ); add ( x + 5, 0, x1, y1, x2, y2 ); add ( x + 5, 5, x1, y1, x2, y2 ); } for ( y = 10; y < y2 + 10; y += 10 ) { //printf ( "2/n" ); add ( 0, y, x1, y1, x2, y2 ); add ( 5, y, x1, y1, x2, y2 ); add ( 0, y + 5, x1, y1, x2, y2 ); add ( 5, y + 5, x1, y1, x2, y2 ); } if ( yy >= 10 ) { //printf ( "3/n" ); for ( x = 10; x < x2 + 10; x += 10 ) { add ( x, yy, x1, y1, x2, y2 ); add ( x, yy + 5, x1, y1, x2, y2 ); add ( x + 5, yy + 5, x1, y1, x2, y2 ); add ( x + 5, yy, x1, y1, x2, y2 ); } } if ( xx >= 10 ) { //printf ( "4/n" ); for ( y = 10; y < yy; y += 10 ) { add ( xx, y, x1, y1, x2, y2 ); add ( xx, y + 5, x1, y1, x2, y2 ); add ( xx + 5, y + 5, x1, y1, x2, y2 ); add ( xx + 5, y, x1, y1, x2, y2 ); } } //printf ( "%lld %lld %lld %lld %lld/n", a[1], a[2], a[3], a[4], a[5] ); long long ret = a[5]; while ( a[4] > 0 ) a[4] --, a[1] --, ret ++; while ( a[3] > 0 ) { ret ++, a[3] --; if ( a[2] > 0 ) a[2] --; else if ( a[1] >= 2 ) a[1] -= 2; else a[1] = 0; } while ( a[2] > 0 ) { ret ++, a[2] --; if ( a[2] > 0 ) { a[2] --; if ( a[1] > 0 ) a[1] --; } else if ( a[1] >= 3 ) a[1] -= 3; else a[1] = 0; } while ( a[1] > 0 ) ret ++, a[1] -= 5; return ret; } }

 

D1 1000p  NowhereLand

 

网络流。这个题的建图还不甚清楚，好像是把所有已有的点和源点相连无穷大，可以建立公司的点和汇点相连无穷大，边与边相连1。

 

 已有点设为黑点，不能建立公司的点设为白点，剩下的点黑白染色，求黑白集合间的最小割。

 

class NowhereLand { int guard[60][60]; int agency[60][60]; int a[60][60]; int prev[60]; int dist[60]; int src, dst; int fflow ( int s, int d, int v ) { queue < int > Q; memset ( dist, 0, sizeof ( dist ) ); Q.push ( s ); dist[s] = 1000000; while ( !Q.empty () && dist[d] == 0 ) { int cur = Q.front (); Q.pop (); printf ( "%d/n", cur ); for ( int i = 0; i < v; i ++ ) { if ( a[cur][i] == 0 ) continue; if ( dist[i] != 0 ) continue; dist[i] = MIN ( dist[cur], a[cur][i] ); prev[i] = cur; Q.push ( i ); } } printf ( "/n%d/n", dist[d] ); return dist[d]; } public: int placeGuards(vector <string> cities, int k, vector <string> guards, vector <string> agencies) { int i, j, n = cities.size (); memset ( guard, 0, sizeof ( guard ) ); memset ( agency, 0, sizeof ( agency ) ); for ( i = 0; i < guards.size (); i ++ ) { istringstream in ( guards[i] ); int x; while ( in >> x ) guard[i][x] = 1; } for ( i = 0; i < agencies.size (); i ++ ) { istringstream in ( agencies[i] ); int x; while ( in >> x ) agency[i][x] = 1; } src = n, dst = n + 1; int flow = 0; int iter; for ( iter = 0; iter < n; iter ++ ) { memset ( a, 0, sizeof ( a ) ); for ( i = 0; i < n; i ++ ) for ( j = 0; j < n; j ++ ) if ( cities[i][j] == '1' ) a[i][j] = 1; for ( i = 0; i < n; i ++ ) if ( !agency[i][iter] ) a[i][dst] = 10000; for ( i = 0; i < n; i ++ ) if ( guard[i][iter] ) a[src][i] = 10000; memset ( prev, 0xff, sizeof ( prev ) ); for ( int f; f = fflow ( src, dst, n + 2 ) > 0; ) { flow += f; for ( int b = dst; b != src; b = prev[b] ) { a[prev[b]][b] -= f; a[b][prev[b]] += f; } } } return flow; } }

 

 

 改进版最大流模板，适用于稀疏矩阵：

 

namespace MF { const int MAXN = 10001; const int MAXM = 30001; const int INF = 1 << 20; vector < PII > e[MAXN]; int flow[MAXM], dist[MAXN], prev[MAXN], edge[MAXN]; int N; int len = 0; int src, dst; queue < int > Q; void clear ( int _N ) { N = _N, len = 0; for ( int i = 0; i < N; i ++ ) e[i].clear (); } void addEdge ( int i, int j, int c1, int c2 ) { //printf ( "%d %d %d %d/n", i, j, c1, c2 ); e[i].push_back ( MP ( j, len ) ); flow[len ++] = c1; e[j].push_back ( MP ( i, len ) ); flow[len ++] = c2; } int bfs () { for ( int i = 0; i < N; i ++ ) dist[i] = 0; dist[src] = INF; Q.push ( src ); while ( !Q.empty () ) { int cur = Q.front (); Q.pop (); for ( int i = 0; i < e[cur].size (); i ++ ) { int v = e[cur][i].first, j = e[cur][i].second; if ( dist[v] || flow[j] == 0 ) continue; dist[v] = min ( dist[cur], flow[j] ); prev[v] = cur, edge[v] = j; Q.push ( v ); } } return dist[dst]; } int maxFlow () { int ret = 0; for ( int f; ( f = bfs () ) > 0; ) { ret += f; for ( int v = dst; v != src; v = prev[v] ) { flow[edge[v]] -= f; int j = ( edge[v] & 1 ) ? edge[v] - 1 : edge[v] + 1; flow[j] += f; } } return ret; } }