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o o o o o o o o o o o o o o o o o o /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F\ /F/ \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \
o o o | o o o o | o o o o o o | o o o o o /F\ /F\ /F\ | /F\ /F\ /F\ /F\ | /F\ /F\ /F\ /F\ /F\ /F\ | /F\ /F\ /F\ /F\ /F/ \ / \ / \ | / \ / \ / \ / \ | / \ / \ / \ / \ / \ / \ | / \ / \ / \ / \ / \
, b0 = 0 and 1 <= k <= M, if there are M groups in total. Note that M can equal to 1.2 5 2 1 4 3 2 5 5 2 5 4 3 2 1
Case #1: 31 Case #2: No solution
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define REP(_,a,b) for(int _=(a); _<=(b);++_)
#define sz(s) (int)((s).size())
typedef long long ll;
const int maxn = 1e5+10;
int n,l;
ll dp[maxn];
struct Num{
ll h;
int idx;
Num(ll h = 0,int idx = 0):h(h),idx(idx){}
friend bool operator < (Num a,Num b){
if(a.h!=b.h) return a.h < b.h;
else return a.idx > b.idx;
}
};
vector<Num> vN;
struct node{
int lson,rson;
ll maxx;
int mid(){
return (lson+rson)>>1;
}
}tree[maxn*4];
void pushUp(int rt){
tree[rt].maxx = max(tree[rt<<1].maxx,tree[rt<<1|1].maxx);
}
void build(int L,int R,int rt){
tree[rt].lson = L;
tree[rt].rson = R;
tree[rt].maxx = -1;
if(L==R){
return;
}
int mid = tree[rt].mid();
build(L,mid,rt<<1);
build(mid+1,R,rt<<1|1);
}
void init(){
vN.clear();
memset(dp,-1,sizeof dp);
}
void update(int pos,int l,int r,int rt,ll x){
if(l==r){
tree[rt].maxx = x;
return;
}
int mid = tree[rt].mid();
if(pos<=mid){
update(pos,l,mid,rt<<1,x);
}else{
update(pos,mid+1,r,rt<<1|1,x);
}
pushUp(rt);
}
ll query(int L,int R,int l,int r,int rt){
if(L <=l && R >= r){
return tree[rt].maxx;
}
int mid = tree[rt].mid();
ll ret;
bool flag = false;
if(L <= mid){
ret = query(L,R,l,mid,rt<<1);
flag = true;
}
if(R > mid){
if(flag){
ret = max(ret,query(L,R,mid+1,r,rt<<1|1));
}else{
ret = query(L,R,mid+1,r,rt<<1|1);
}
}
return ret;
}
void input(){
scanf("%d%d",&n,&l);
REP(_,1,n){
ll h;
scanf("%I64d",&h);
vN.push_back(Num(h,_));
}
sort(vN.begin(),vN.end());
build(0,n,1);
}
void solve(){
update(0,0,n,1,0);
REP(_,0,sz(vN)-1) {
int ni = vN[_].idx;
ll nh = vN[_].h;
ll tm = query(max(ni-l,0),ni-1,0,n,1);
if(tm>=0){
dp[ni] = tm+nh*nh;
update(ni,0,n,1,dp[ni]-nh);
}
if(ni==n) break;
}
if(dp[n]<=0){
printf("No solution\n");
}else{
printf("%I64d\n",dp[n]);
}
}
int main(){
int ncase,T=1;
cin >> ncase;
while(ncase--){
init();
input();
printf("Case #%d: ",T++);
solve();
}
return 0;
} HDU4719-Oh My Holy FFF(DP线段树优化)
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原文地址:http://www.cnblogs.com/mengfanrong/p/5173851.html