http://cogs.pro/cogs/problem/problem.php?pid=1752?(題目鏈接)

題意

　　給出$n*n$的棋盤，單點修改，矩陣查詢。

Solution

　　離線以后CDQ分治。每一層按照$Y$排序，然后詢問用前綴和拆成$4$個，樹狀數組維護一下就可以了。

細節

　　？

代碼

// cogs1752 #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> #include<cmath> #define LL long long #define inf (1ll<<30) #define Pi acos(-1.0) #define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout) using namespace std;const int maxn=200010,maxd=2000010; int c[maxd],ans[maxn],n,m; struct data {int x,y,k,id,op;}q[maxn<<2],nq[maxn<<2];int lowbit(int x) {return x&-x; } void add(int x,int val) {for (int i=x;i<=n;i+=lowbit(i)) c[i]+=val; } int query(int x) {int res=0;for (int i=x;i;i-=lowbit(i)) res+=c[i];return res; } void solve(int l,int r) {if (l==r) return;int mid=(l+r)>>1,l1=l,l2=mid+1;for (int i=l;i<=r;i++) q[i].id<=mid ? nq[l1++]=q[i] : nq[l2++]=q[i];for (int i=l;i<=r;i++) q[i]=nq[i];solve(l,mid);solve(mid+1,r);for (int i=l,j=mid+1,k=l;i<=mid || j<=r;) {if (j>r || (i<=mid && q[i].y<=q[j].y)) {if (q[i].k>0) add(q[i].x,q[i].k);nq[k++]=q[i++];}else {if (q[j].k<0) ans[-q[j].k]+=q[j].op*query(q[j].x);nq[k++]=q[j++];}}for (int i=l;i<=mid;i++) if (q[i].k>0) add(q[i].x,-q[i].k);for (int i=l;i<=r;i++) q[i]=nq[i]; } int main() {free("mokia");int T,tot=0;while (scanf("%d",&T)!=EOF) {if (T==0) scanf("%d",&n);if (T==1) {int x,y,k;scanf("%d%d%d",&x,&y,&k);m++,q[m]=(data){x,y,k,m,0};}if (T==2) {int x1,y1,x2,y2;tot++;scanf("%d%d%d%d",&x1,&y1,&x2,&y2);m++,q[m]=(data){x1-1,y1-1,-tot,m,1};m++,q[m]=(data){x1-1,y2,-tot,m,-1};m++,q[m]=(data){x2,y1-1,-tot,m,-1};m++,q[m]=(data){x2,y2,-tot,m,1};}if (T==3) break;}solve(1,m);for (int i=1;i<=tot;i++) printf("%d\n",ans[i]);return 0; }

?

轉載于:https://www.cnblogs.com/MashiroSky/p/6516950.html

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