A The beautiful values of the palace
Knowledge Point: Analog, Fenwick tree + prefix and two-dimensional discrete +
- The calculated coordinate value (as a first center, and then subtracting n * n, in turn easier to count)
- The special points to queue
- Reads the inquiry, the inquiry \ ((x_1, Y_1, x_2, y_2) \) , if \ (mp [x] [y ] \) as a prefix and a two-dimensional, the answer is \ (mp [x_2] [y_2 ] -mp [x_2] [y_1-1] -mp [x_1-1] [y_2] + mp [x_1-1] [y_1-1] \)
- All query prefix requires a two-dimensional coordinates obtained also queued to be sorted according to a first reference x, y of discretization
- After ordering from small to large order of consideration, that is, x is from small to large consideration, every time they meet a new special point on the array is inserted into the tree, interrogated \ ((x, y) \ ) in Fenwick tree query (1, y-1) and can.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1000010;
int T;
int n,m,p;
ll d[N],c[N];
struct node{
int x1,y1,x2,y2;
};
vector<node> qu;//查询
struct Node{
int x,y;
int type;//type为0表示添加新点
}a[N];
bool cmp(Node a,Node b){
if(a.x == b.x){
if(a.y == b.y)return a.type == 0;
return a.y < b.y;
}
return a.x < b.x;
}
vector<int> v;//离散y
void add(int x,int y){
for(;x<N;x+=x&-x)c[x] += y;
}
ll ask(int x){
ll res = 0;
for(;x;x-=x&-x)res += c[x];
return res;
}
map<int,map<int,int>> mp;
int id[N];
ll get(int x,int y){
int cx = n / 2 + 1;
int cy = n / 2 + 1;//(cx,cy)即中心坐标
int k = max(abs(x-cx),abs(y-cy));//k表示在第几圈
if(k == 0)return 1ll * n * n;
//printf("\n(%d,%d) k : %d\n",x,y,k);
ll res = d[k-1];//圈内有多少个
//分4个case计算
if(y-cy == k && x < cx + k){//在上层
res += cx + k - x;
}else if(cx - x == k && y < cy + k){//在左侧
res += k * 2 + cy + k - y;
}else if(cy - y == k && x > cx - k){//在下层
res += k * 4 + x - (cx - k);
}else if(x - cx == k){//在右侧
res += k * 6 + y - (cy - k);
}
res = 1ll * n * n - res;//最后倒过来，因为上面是按照中心为1算的
return res + 1;
}
int calc(ll x){
int res= 0;
while(x){res += x % 10;x /= 10;}
return res;
}
int main(){
scanf("%d",&T);
d[0] = 1;d[1] = 8;
for(int i=2;i<N;i++)d[i] = d[i-1] + 8;
for(int i=1;i<N;i++)d[i] += d[i-1];//计算i圈以内有多少个点
while(T--){
mp.clear();v.clear();qu.clear();
memset(c,0,sizeof c);
scanf("%d%d%d",&n,&m,&p);
for(int i=1;i<=m;i++){
int x,y;
scanf("%d%d",&x,&y);
a[i] = {x,y,0};
}
for(int i=1;i<=p;i++){
int x1,x2,y1,y2;
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
qu.push_back({x1,y1,x2,y2});
a[++m] = {x1-1,y1-1,1};
a[++m] = {x2,y2,1};
a[++m] = {x1-1,y2,1};
a[++m] = {x2,y1-1,1};
}
for(int i=1;i<=m;i++){
v.push_back(a[i].y);
}
sort(v.begin(),v.end());
v.erase(unique(v.begin(),v.end()),v.end());
sort(a+1,a+m+1,cmp);
for(int i=1;i<=m;i++){
id[i] = lower_bound(v.begin(),v.end(),a[i].y) - v.begin() + 1;
}
for(int i=1;i<=m;i++){
if(a[i].type == 0)add(id[i],calc(get(a[i].x,a[i].y)));
else if(a[i].type == 1)mp[a[i].x][a[i].y] = ask(id[i]);
}
for(int i=0;i<p;i++){
int x1 = qu[i].x1,x2 = qu[i].x2,y1 = qu[i].y1,y2 = qu[i].y2;
ll res = mp[x2][y2] + mp[x1-1][y1-1] - mp[x1-1][y2] - mp[x2][y1-1];
printf("%lld\n",res);
}
}
return 0;
}D Robots
Set \ (D [u] \) is the desired number of days to u n, then there are:
\(d[u] = d[u] * {1\over outdeg[u]} + {1\over outdeg[u]} * \sum d[v] + 1\)
Set \ (F [u] \) of u n to the desired consumption, it has:
\(f[u] = {1\over outdeg[u]} *(f[u] + d[u]) + {1\over outdeg[u]} * \sum (f[v]+d[v]) + 1\)
Because you this day, every day after will have contributed an additional 1 u this point in the past, so to add d [u] and \ (\ sum d [v] \)
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6+10;
int head[N],ver[N],nxt[N],tot,deg[N],sz[N];
int T,n,m;
void add(int x,int y){
ver[++tot] = y;nxt[tot] = head[x];head[x] = tot;
}
double f[N],d[N];
int main(){
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
tot = 0;
for(int i=1;i<=n;i++){
head[i] = 0;f[i] = d[i] = 0;deg[i] = 0;sz[i] = 0;
}
for(int i=1;i<=m;i++){
int x,y;scanf("%d%d",&x,&y);
add(y,x);
deg[x]++;sz[x]++;
}
queue<int> q;
q.push(n);
while(q.size()){
int x = q.front();q.pop();
for(int i=head[x];i;i=nxt[i]){
int y = ver[i];
f[y] += f[x];
d[y] += d[x];
if(-- deg[y] == 0){
//if刚进来时，f[y] 和 d[y] 为 累加和
f[y] = ((d[y]+sz[y]+1)/sz[y]+f[y]+d[y]+sz[y]+1)/sz[y];
d[y] = (d[y] + sz[y] + 1)/sz[y];
q.push(y);
}
}
}
printf("%.2f\n",f[1]);
}
return 0;
}F Greedy Sequence
Weight maximum number of segment tree maintenance occurred, i.e., [l,r]represents the number of all the current, the value in the \ ([l, r] \ ) the maximum number of the number range
Slide interval maintenance segment tree for every query, [1,x-1]the maximum number that appears if there is no answer 0
Finally, remember to search
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int T;
const int N = 100010;
int a[N],n,k,b[N],d[N];
struct Seg{
int l,r;
int mx;
}t[4*N];
void build(int p,int l,int r){
t[p].l = l;t[p].r = r;
if(l == r){
t[p].mx = 0;return;
}
int mid = l + r >> 1;
build(p*2,l,mid);
build(p*2+1,mid+1,r);
t[p].mx = 0;
}
void upd(int p,int x,int add){
if(t[p].l == t[p].r && t[p].l == x){
t[p].mx = add == 1 ? x : 0;
return;
}
int mid = t[p].l + t[p].r >> 1;
if(mid >= x)upd(p*2,x,add);
if(mid < x)upd(p*2+1,x,add);
t[p].mx = max(t[p*2].mx,t[p*2+1].mx);
}
int query(int p,int l,int r){
if(t[p].l >= l && t[p].r <= r){
return t[p].mx;
}
int mid = t[p].l + t[p].r >> 1;
int res = 0;
if(mid >= l)res = max(res,query(p*2,l,r));
if(mid < r)res = max(res,query(p*2+1,l,r));
return res;
}
ll get(int x){
if(b[x] == 0)return d[x] = 1;
if(d[x])return d[x];
d[x] = get(b[x]) + 1;
return d[x];
}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
for(int i=1;i<=n;i++)d[i] = 0;
int l = 1,r=min(1+k,n);
build(1,1,n);
for(int i=l;i<=r;i++){
upd(1,a[i],1);
}
for(int i=1;i<=n;i++){
b[a[i]] = query(1,1,a[i]-1);
if(i-l>=k){
upd(1,a[l],0);
l++;
}
if(r<n){
upd(1,a[++r],1);
}
}
ll res = 0;
for(int i=1;i<=n;i++)res += get(a[i]);
for(int i=1;i<=n;i++){
if(i>1)printf(" ");
printf("%d",get(i));
}
puts("");
}
return 0;
}H. Holy Grail
Every time I go the shortest path to return the answer is positive then add the negative side right, otherwise add positive side (be careful not to use Dijkstra)
#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
typedef long long ll;
const ll INF = 1e18;
int head[N],ver[N],Nxt[N],tot;
int v[N],n,m;
ll edge[N],d[N];
void add(int x,int y,ll z){
ver[++tot] = y;edge[tot] = z;Nxt[tot] = head[x];head[x] = tot;
}
ll dij(int s,int t){
queue<int> q;
for(int i = 0; i <= n + 5; i++){
d[i] = INF;v[i] = 0;
}
d[s] = 0; q.push(s); v[s] = 1;
while(q.size())
{
int x = q.front(); q.pop();
v[x] = 0;
for(int i = head[x]; i; i = Nxt[i]){
int y = ver[i];
if(d[y] > d[x] + edge[i])
{
d[y] = d[x] + edge[i];
if(v[y] == 0){
v[y] = 1;
q.push(y);
}
}
}
}
return d[t];
}
int main(){
int T;scanf("%d",&T);
for(int o=0;o<T;o++){
scanf("%d%d",&n,&m);
tot = 0;
for(int i=1;i<=n;i++)head[i] = 0;
for(int i=1;i<=m;i++){
int x,y;
ll z;
scanf("%d%d%lld",&x,&y,&z);
x++;y++;
add(x,y,z);
}
for(int j=1;j<=6;j++){
int s,t;
scanf("%d%d",&s,&t);
s++;t++;
ll dis = dij(t,s);
printf("%lld\n",-dis);
add(s,t,-dis);
}
}
return 0;
}