Problem Description

There are?nn?towns in Byteland, labeled by?1,2,\dots,n1,2,…,n. The?ii-th town's location is?(x_i,y_i)(xi?,yi?). Little Q got a taxi VIP card, he can use the VIP card to cut down the taxi fare. Formally, assume Little Q is at?(x',y')(x′,y′), if he calls a taxi to drive him to the?kk-th town, the VIP card will reduce?\min(|x'-x_k|+|y'-y_k|,w_k)min(∣x′?xk?∣+∣y′?yk?∣,wk?)?dollars.

Little Q wants to make full use of his VIP card. He will give you?qq?queries, in each query you will be given his location, and you need to choose a town such that the VIP card will reduce the most taxi fare.

Input

The first line contains a single integer?TT?(1 \leq T \leq 1001≤T≤100), the number of test cases. For each test case:

The first line contains two integers?nn?and?qq?(1 \leq n,q \leq 100,0001≤n,q≤100,000), denoting the number of towns and the number of queries.

Each of the following?nn?lines contains three integers?x_ixi?,?y_iyi??and?w_iwi??(1 \leq x_i,y_i,w_i \leq 10^91≤xi?,yi?,wi?≤109), describing a town.

Each of the following?qq?lines contains two integers?x'x′?and?y'y′?(1 \leq x',y' \leq 10^91≤x′,y′≤109), describing a query.

It is guaranteed that the sum of all?nn?is at most?500,000500,000, and the sum of all?qq?is at most?500,000500,000.

Output

For each query, print a single line containing an integer, denoting the maximum possible reduced taxi fare.

#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; const int N=1e5+5,INF=2e9; struct node{int x,y,w;bool operator < (const node &b){return w<b.w;} }p[N]; int a[N],b[N],c[N],d[N]; int xx,yy; int ans=0; bool pd(int mid) {int mxv=0;mxv=max(mxv,xx+yy+a[mid]);mxv=max(mxv,xx-yy+b[mid]);mxv=max(mxv,-xx+yy+c[mid]);mxv=max(mxv,-xx-yy+d[mid]);ans=max(ans,min(mxv,p[mid].w));return mxv>=p[mid].w; } void solve() {int n,q;scanf("%d%d",&n,&q);for(int i=1;i<=n;i++){int x,y,w;scanf("%d%d%d",&x,&y,&w);p[i]={x,y,w};}sort(p+1,p+1+n);a[n+1]=b[n+1]=c[n+1]=d[n+1]=-INF;for(int i=n;i>=1;i--){a[i]=max(a[i+1],-p[i].x-p[i].y);b[i]=max(b[i+1],-p[i].x+p[i].y);c[i]=max(c[i+1],p[i].x-p[i].y);d[i]=max(d[i+1],p[i].x+p[i].y);}while(q--){scanf("%d%d",&xx,&yy);ans=0;int l=1,r=n;while(l<=r){int m=(l+r)/2;if(pd(m)) l=m+1;else r=m-1;}printf("%d\n",ans);}} int main() {int t;scanf("%d",&t);while(t--){solve();} }

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