liaoliao的四连做第二弹
1.bzoj3211: 花神游历各国
由于$10^9$以内的数最多只会被开方$10$次,所以我们可以用线段树维护然后剪枝..
#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <cmath> #define LL long long using namespace std; const LL Maxn = 100010; LL p[Maxn][31], nochange[Maxn][31]; LL sum[Maxn*4], la[Maxn*4]; bool isone[Maxn*4]; LL n, m; void bulid_tree ( LL now, LL L, LL R ){ if ( L < R ){ LL mid = ( L + R ) >> 1, lc = now*2, rc = now*2+1; bulid_tree ( lc, L, mid ); bulid_tree ( rc, mid+1, R ); sum[now] = sum[lc]+sum[rc]; la[now] = 0; if ( isone[lc] && isone[rc] ) isone[now] = true; } else { sum[now] = p[L][0]-p[L-1][0]; la[now] = 0; if ( sum[now] <= 1 ) isone[now] = true; } } void push_down ( LL now, LL L, LL R ){ LL mid = ( L + R ) >> 1, lc = now*2, rc = now*2+1; if ( la[now] >= 0 ){ la[lc] = la[rc] = la[now]; sum[lc] = p[mid][la[lc]]-p[L-1][la[lc]]; if ( nochange[mid][la[lc]]-nochange[L-1][la[lc]] == mid-L+1 ) isone[lc] = true; sum[rc] = p[R][la[rc]]-p[mid][la[rc]]; if ( nochange[R][la[rc]]-nochange[mid][la[rc]] == R-mid ) isone[rc] = true; } } void change ( LL now, LL L, LL R, LL l, LL r ){ if ( isone[now] ) return; if ( L == l && R == r && la[now] != -1 ){ la[now] ++; sum[now] = p[R][la[now]]-p[L-1][la[now]]; if ( nochange[R][la[now]]-nochange[L-1][la[now]] == (R-L+1) ) isone[now] = true; return; } push_down ( now, L, R ); LL mid = ( L + R ) >> 1, lc = now*2, rc = now*2+1; if ( r <= mid ) change ( lc, L, mid, l, r ); else if ( l > mid ) change ( rc, mid+1, R, l, r ); else change ( lc, L, mid, l, mid ), change ( rc, mid+1, R, mid+1, r ); sum[now] = sum[lc]+sum[rc]; if ( la[lc] == la[rc] ) la[now] = la[lc]; else la[now] = -1; if ( isone[lc] && isone[rc] ) isone[now] = true; } LL query ( LL now, LL L, LL R, LL l, LL r ){ if ( L == l && R == r ) return sum[now]; push_down ( now, L, R ); LL mid = ( L + R ) >> 1, lc = now*2, rc = now*2+1; if ( r <= mid ) return query ( lc, L, mid, l, r ); else if ( l > mid ) return query ( rc, mid+1, R, l, r ); else return query ( lc, L, mid, l, mid ) + query ( rc, mid+1, R, mid+1, r ); } int main (){ LL i, j, k; scanf ( "%lld", &n ); for ( i = 1; i <= n; i ++ ){ scanf ( "%lld", &p[i][0] ); if ( p[i][0] <= 1 ) nochange[i][0] = 1; for ( j = 1; j <= 10; j ++ ){ p[i][j] = (LL)sqrt(p[i][j-1]); if ( p[i][j] <= 1 ) nochange[i][j] = 1; } } for ( j = 0; j <= 10; j ++ ){ for ( i = 1; i <= n; i ++ ) p[i][j] += p[i-1][j], nochange[i][j] += nochange[i-1][j]; } bulid_tree ( 1, 1, n ); scanf ( "%lld", &m ); for ( i = 1; i <= m; i ++ ){ LL fl, l, r; scanf ( "%lld%lld%lld", &fl, &l, &r ); if ( fl == 1 ){ printf ( "%lld\n", query ( 1, 1, n, l, r ) ); } else { change ( 1, 1, n, l, r ); } } return 0; }
2.bzoj4240: 有趣的家庭菜园
考虑一个贪心策略,从小到大移动,小的数肯定放两边(哪边近放哪边)
那么对于某个数能够影响他的也就只有比他大的数了
然后就从大到小插入然后树状数组判断一下就好了..
#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #define LL long long using namespace std; const LL Maxn = 300010; struct node { int x, pos; }list[Maxn]; bool cmp ( node x, node y ){ return x.x > y.x; } LL sum[Maxn], n; LL lowbit ( LL x ){ return x & (-x); } void add ( LL x ){ while ( x <= n ){ sum[x] ++; x += lowbit (x); } } LL query ( LL x ){ LL ret = 0; while ( x > 0 ){ ret += sum[x]; x -= lowbit (x); } return ret; } LL _min ( LL x, LL y ){ return x < y ? x : y; } int main (){ LL i, j, k; scanf ( "%lld", &n ); for ( i = 1; i <= n; i ++ ) scanf ( "%lld", &list[i].x ), list[i].pos = i; sort ( list+1, list+n+1, cmp ); memset ( sum, 0, sizeof (sum) ); LL ans = 0; LL num = 1; for ( i = 1; i <= n; i ++ ){ if ( list[i].x != list[i-1].x ){ while ( num < i ) add (list[num].pos), num ++; } LL s = query (list[i].pos); ans += _min ( s, num-s-1 ); } printf ( "%lld\n", ans ); return 0; }
3.bzoj3232: 圈地游戏
二分答案,然后用网络流判断,是一个比较经典的最小割模型..
像这种小数流量的,可以先把这个数扩大$10^6$最后再缩小..
听说这种做法叫做分数规划(雾)
#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <queue> #define LL long long using namespace std; const LL Maxn = 55; const LL Maxm = 55*55; const LL inf = 0x7fffffff; struct node { LL y, next, opp; LL c; }a[Maxm*20]; LL first[Maxm], len; void ins ( LL x, LL y, LL c ){ len ++; LL k1 = len; a[len].y = y; a[len].c = c; a[len].next = first[x]; first[x] = len; len ++; LL k2 = len; a[len].y = x; a[len].c = 0; a[len].next = first[y]; first[y] = len; a[k1].opp = k2; a[k2].opp = k1; } LL n, m; LL st, ed, h[Maxm]; LL getnum ( LL x, LL y ){ return (x-1)*m+y; } LL map[Maxn][Maxn]; LL row[Maxn][Maxn], col[Maxn][Maxn]; LL _min ( LL x, LL y ){ return x < y ? x : y; } LL dfs ( LL x, LL flow ){ if ( x == ed ) return flow; LL delta = 0; for ( LL k = first[x]; k; k = a[k].next ){ LL y = a[k].y; if ( h[y] == h[x]+1 && a[k].c > 0 && flow-delta > 0 ){ LL minf = dfs ( y, _min ( a[k].c, flow-delta ) ); delta += minf; a[k].c -= minf; a[a[k].opp].c += minf; } } if ( delta == 0 ) h[x] = -1; return delta; } bool bfs (){ queue <LL> q; memset ( h, -1, sizeof (h) ); q.push (st); h[st] = 0; while ( !q.empty () ){ LL x = q.front (); q.pop (); for ( LL k = first[x]; k; k = a[k].next ){ LL y = a[k].y; if ( h[y] == -1 && a[k].c > 0 ){ h[y] = h[x]+1; q.push (y); } } } return h[ed] > 0; } int main (){ LL i, j, k; scanf ( "%lld%lld", &n, &m ); LL sum = 0; for ( i = 1; i <= n; i ++ ) for ( j = 1; j <= m; j ++ ){ scanf ( "%lld", &map[i][j] ); sum += map[i][j]; } sum *= 1e6; for ( i = 1; i <= n+1; i ++ ) for ( j = 1; j <= m; j ++ ) scanf ( "%lld", &row[i][j] ); for ( i = 1; i <= n; i ++ ) for ( j = 1; j <= m+1; j ++ ) scanf ( "%lld", &col[i][j] ); LL l = 0, r = 100*1e6, ret; st = 0; ed = n*m+1; while ( l <= r ){ LL mid = (l+r)>>1; len = 0; memset ( first, 0, sizeof (first) ); for ( i = 1; i <= n; i ++ ){ for ( j = 1; j <= m; j ++ ){ LL x = getnum (i,j); if ( i == 1 ) ins ( st, x, row[1][j]*mid ); if ( i == n ) ins ( st, x, row[n+1][j]*mid ); if ( j == 1 ) ins ( st, x, col[i][1]*mid ); if ( j == m ) ins ( st, x, col[i][m+1]*mid ); ins ( x, ed, map[i][j]*1e6 ); LL ii = i, jj = j+1; if ( ii >= 1 && ii <= n && jj >= 1 && jj <= m ){ ins ( x, getnum(ii,jj), col[i][j+1]*mid ); ins ( getnum(ii,jj), x, col[i][j+1]*mid ); } ii = i+1; jj = j; if ( ii >= 1 && ii <= n && jj >= 1 && jj <= m ){ ins ( x, getnum(ii,jj), row[i+1][j]*mid ); ins ( getnum(ii,jj), x, row[i+1][j]*mid ); } } } LL delta = 0; while ( bfs () ){ delta += dfs ( st, inf*1e6 ); } if ( sum-delta > 0 ){ ret = mid; l = mid+1; } else r = mid-1; } printf ( "%.3lf\n", (double)ret/1e6 ); return 0; }
4.bzoj3162: 独钓寒江雪
先找重心,因为重心肯定是刚好对应的(如果有两个重心就新建一个重心练到这两个点然后再乱搞..)
那么剩下的肯定只有是在交换子树下才有可能同形异构..
这个东西就是一个可重复排列..
代码先挖个坑..
作者:Ra1nbow
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