传送门:CodeForces-1216F
思路分析
DP方程不难想,转移过程是区间最值,所以可以用线段树优化
AC代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
bool IOerror=0;
inline char nc() {
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if(p1==pend) {
p1=buf;
pend=buf+fread(buf,1,BUF_SIZE,stdin);
if(pend==p1) {
IOerror=1;
return -1;
}
}
return *p1++;
}
inline bool blank(char ch) {
return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
}
template<class T> inline bool read(T &x) {
bool sign=0;
char ch=nc();
x=0;
for(; blank(ch); ch=nc());
if(IOerror)return false;
if(ch=='-')sign=1,ch=nc();
for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
if(sign)x=-x;
return true;
}
template<class T,class... U>bool read(T& h,U&... t) {
return read(h)&&read(t...);
}
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define int long long
const int INF = 1e18;
const int N = 2e5+10;
struct node{
struct Node{
int l,r,mid,mi;
}tree[N<<2];
#define ls x<<1
#define rs x<<1|1
void pp(int x){
tree[x].mi=min(tree[ls].mi,tree[rs].mi);
}
void build(int x,int l,int r){
tree[x]={l,r,l+r>>1,INF};
if(l==r) return ;
int mid=tree[x].mid;
build(ls,l,mid);
build(rs,mid+1,r);
pp(x);
}
void modify(int x,int pos,int k){
if(tree[x].l==pos && tree[x].r==pos){
tree[x].mi=min(tree[x].mi,k);
return ;
}
int mid=tree[x].mid;
if(pos<=mid) modify(ls,pos,k);
else modify(rs,pos,k);
pp(x);
}
int query(int x,int l,int r){
if(l<=tree[x].l && tree[x].r<=r ){
return tree[x].mi;
}
int ans=INF;
int mid=tree[x].mid;
if(l<=mid) ans=min(ans,query(ls,l,r));
if(r>mid) ans=min(ans,query(rs,l,r));
return ans;
}
#undef ls
#undef rs
}t0,t1;
int dp[N][2];
char s[N];
signed main() {
#ifdef xiaofan
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
int n,k;
cin>>n>>k;
cin>>s+1;
for(int i=1;i<=n;i++) dp[i][0]=dp[i][1]=1e18;
dp[0][0]=0;
t0.build(1,0,n);
t1.build(1,0,n);
t0.modify(1,0,0);
for(int i=1;i<=n;i++){
dp[i][0]=min(dp[i-1][0]+i,t1.query(1,max(0LL,i-k),i-1));
if(s[i]=='1') dp[i][1]=min(t0.query(1,max(0LL,i-k-1),i-1),t1.query(1,max(0LL,i-k-1),i-1))+i;
t0.modify(1,i,dp[i][0]);
t1.modify(1,i,dp[i][1]);
}
cout<<min(dp[n][0],dp[n][1])<<endl;
return 0;
}