[莫队算法] - cool's Blog
[莫队算法]
最近又看了一下莫队，发现我以前完全理解错了莫队算法。
莫队算法适用于无修（带修不会啊QAQ@lbn187）的区间询问且若已知[L,R]的答案能够以较快的速度求出[L-1,R],[L+1,R],[L,R-1],[L,R+1],那么就可以使用莫队。
莫队是将所有区间按照一定顺序排序，使得这个区间移动次数较少。
最优的方法就是曼哈顿距离的最小生成树。但写起来太麻烦。
用这种方法排序是比较优秀的：如果L1/blk=L2/blk，那么就按R排序，否则按L排序。
经典例题小Z的袜子（叫你每天代码抄抄）
[51nod1290]询问区间内多少对数差的绝对值<=k，k为全局给定。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 | #include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define rep(i,a,b) for (int i = a;i <= b;i++)#define N 50010#define M 150010#define ll long longusing namespace std;struct zhl{ int l,r,id; zhl(){} zhl(int L,int R,int ID):l(L), r(R), id(ID){}}eg[N];int l,r,n,k,q,pos[N],blk,len,gg[M],a[N],b[N],c[N],x,y;ll Ans,ans[N];int cmp(zhl x,zhl y){ return pos[x.l]^pos[y.l]?x.l < y.l:x.r < y.r;}int g[M];void update(int x,int y){ for (;x <= len;x += x&-x) g[x] += y;}int query(int x){ int y = 0; for (;x;x -= x&-x) y += g[x]; return y;}void solve(){ l = r = 1, Ans = 0; update(a[1],1); rep(i,1,q) { int L = eg[i].l,R = eg[i].r; for (;l < L;l++) update(a[l],-1),Ans -= query(c[l])-query(b[l]-1); for (;l > L;update(a[l],1)) l--,Ans += query(c[l])-query(b[l]-1); for (;r < R;update(a[r],1)) r++,Ans += query(c[r])-query(b[r]-1); for (;r > R;r--) update(a[r],-1),Ans -= query(c[r])-query(b[r]-1); ans[eg[i].id] = Ans; }} int main(){ scanf("%d%d%d",&n,&k,&q); blk = sqrt(n); rep(i,1,n) { scanf("%d",&a[i]); gg[++len] = a[i]; gg[++len] = a[i]+k; gg[++len] = a[i]-k; } sort(gg+1,gg+len+1); len = unique(gg+1,gg+len+1)-gg-1; rep(i,1,n) { b[i] = lower_bound(gg+1,gg+len+1,a[i]-k)-gg; c[i] = lower_bound(gg+1,gg+len+1,a[i]+k)-gg; a[i] = lower_bound(gg+1,gg+len+1,a[i])-gg; pos[i] = (i-1)/blk; } rep(i,1,q) { scanf("%d%d",&x,&y); eg[i] = zhl(x+1,y+1,i); } sort(eg+1,eg+q+1,cmp); solve(); rep(i,1,q) printf("%lld\n",ans[i]); return 0;} |
[bzoj4540]求一段区间的所有子区间的最小值的和。
考虑加入一个数的贡献，令p为[L,R]的最小值。
那么（p-L+1）*a[p]是其中一部分的贡献，还有的贡献在[p+1,R]。
令sl[i]为1~i的答案，那么sl[R]-sl[p]就是[p+1,R]的答案。
为什么呢？因为p是最小值，也就是说[p+1,R]的所有值的sl都是以sl[p]为基础转移的。
还有一点，莫队的时候应该先增加区间，再减小区间，不然有可能会有L>R的情况，会biubiu。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 | #include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <map>#define rep(i,a,b) for (int i = a;i <= b;i++)#define drep(i,a,b) for (int i = a;i >= b;i--)#define ist insert#define ers erase#define clr(a,b) memset(a,b,sizeof a)#define ll long long#define inf 1000000000#define zhl 1000000007template<class T>void Swap(T &a,T &b){ T c = a;a = b;b = c; }template<class T>T max(T& a,T& b){ return a<b?b:a; }template<class T>T min(T& a,T& b){ return a>b?b:a; }using namespace std;#define N 100010int L[N],R[N],stk[N],pos[N],pw[20],n,q,top,blk,Left,Right;ll sl[N],sr[N],a[N],ans[N],now;int lg[N],f[N][20];struct xfy{ int l,r,id;}eg[N];void checkmin(int& x,int y){ if (a[x] > a[y]) x = y;}int getmin(int x,int y){ if (x > y) Swap(x,y); int l = lg[y-x+1],r1 = f[x][l],r2 = f[y-pw[l]+1][l]; return a[r1]<a[r2]?r1:r2;}void updatel(int l,int r,int ty){ int o = getmin(l,r); ll res = a[o]*(r-o+1)+sr[l]-sr[o]; now += res*ty;}void updater(int l,int r,int ty){ int o = getmin(l,r); ll res = a[o]*(o-l+1)+sl[r]-sl[o]; now += res*ty;}int cmp(xfy x,xfy y){ return pos[x.l] == pos[y.l]?x.r<y.r:x.l<y.l;}int main(){ scanf("%d%d",&n,&q); blk = sqrt(n); pw[0] = 1; rep(i,1,18) pw[i] = pw[i-1]+pw[i-1]; rep(i,1,n) scanf("%lld",&a[i]); rep(i,1,n) { for (;top&&a[stk[top]]>=a[i];R[stk[top--]]=i); stk[++top] = i; } for (;top;R[stk[top--]]=n+1); drep(i,n,1) { for (;top&&a[stk[top]]>a[i];L[stk[top--]]=i); stk[++top] = i; } for (;top;L[stk[top--]]=0); rep(i,1,n) sl[i] = sl[L[i]]+a[i]*(i-L[i]); drep(i,n,1) sr[i] = sr[R[i]]+a[i]*(R[i]-i); lg[0] = -1; rep(i,1,n) lg[i] = lg[i>>1]+1,pos[i] = (i-1)/blk,f[i][0] = i; rep(j,1,18) { rep(i,1,n) { f[i][j] = f[i][j-1]; if (i+pw[j-1] <= n) checkmin(f[i][j],f[i+pw[j-1]][j-1]); } } rep(i,1,q) scanf("%d%d",&eg[i].l,&eg[i].r),eg[i].id = i; sort(eg+1,eg+q+1,cmp); Left = Right = 1;now = a[1]; rep(i,1,q) { int l = eg[i].l,r = eg[i].r; for (;Right < r;Right++) updater(Left,Right+1,1); for (;Left > l;Left--) updatel(Left-1,Right,1); for (;Right > r;Right--) updater(Left,Right,-1); for (;Left < l;Left++) updatel(Left,Right,-1); ans[eg[i].id] = now; } rep(i,1,q) printf("%lld\n",ans[i]); return 0;} |
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