Bzoj1835：[ZJOI2010]基站选址

Sol

设\(f[i][j]\)表示钦定\(i\)建基站，建了\(j\)个基站的最小代价
\(f[i][j]=max(f[l][j-1]+\Sigma_{t=l+1}^{i-1}\)不能影响到的村庄的\(w[t])+c[i]\)

二分处理出每个村庄\(p\)左右能影响到它的最远的基站设为\(L[p], R[p]\)
\(l,i\)不能影响到的即\(L[p]>l, R[p]<i\)

枚举\(j\)，预处理出\(j=1\)的情况
线段树
每次把上次的\(f\)重建进入线段树，维护最小值
再枚举\(i\)每次加入\(R[p]\)小于\(i\)的覆盖\(1,L[p]\)

我是做到\(f[n+1]\)，然后做\(k+1\)遍直接输出\(f[n+1]\)的

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e5 + 5);

IL ll Input(){
    RG ll x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, k, d[_], s[_], w[_], c[_], l[_], r[_], first[_], nxt[_];
int mn[_ << 2], tag[_ << 2], f[_];

IL void Build(RG int x, RG int l, RG int r){
    tag[x] = 0;
    if(l == r){
        mn[x] = f[l];
        return;
    }
    RG int mid = (l + r) >> 1;
    Build(x << 1, l, mid), Build(x << 1 | 1, mid + 1, r);
    mn[x] = min(mn[x << 1], mn[x << 1 | 1]);
}

IL void Modify(RG int x, RG int l, RG int r, RG int L, RG int R, RG int v){
    if(L <= l && R >= r){
        mn[x] += v, tag[x] += v;
        return;
    }
    RG int mid = (l + r) >> 1;
    if(L <= mid) Modify(x << 1, l, mid, L, R, v);
    if(R > mid) Modify(x << 1 | 1, mid + 1, r, L, R, v);
    mn[x] = min(mn[x << 1], mn[x << 1 | 1]) + tag[x];
}

IL int Query(RG int x, RG int l, RG int r, RG int L, RG int R){
    if(R < L) return 0;
    if(L <= l && R >= r) return mn[x];
    RG int mid = (l + r) >> 1, ans = 2e9;
    if(L <= mid) ans = Query(x << 1, l, mid, L, R);
    if(R > mid) ans = min(ans, Query(x << 1 | 1, mid + 1, r, L, R));
    return ans + tag[x];
}

int main(RG int argc, RG char *argv[]){
    Fill(first, -1), n = Input(), k = Input();
    for(RG int i = 2; i <= n; ++i) d[i] = Input();
    for(RG int i = 1; i <= n; ++i) c[i] = Input();
    for(RG int i = 1; i <= n; ++i) s[i] = Input();
    for(RG int i = 1; i <= n; ++i) w[i] = Input();
    for(RG int i = 1; i <= n; ++i){
        RG int L = 1, R = i;
        while(L <= R){
            RG int mid = (L + R) >> 1;
            if(d[i] - d[mid] <= s[i]) R = mid - 1, l[i] = mid;
            else L = mid + 1;
        }
        L = i, R = n;
        while(L <= R){
            RG int mid = (L + R) >> 1;
            if(d[mid] - d[i] <= s[i]) L = mid + 1, r[i] = mid;
            else R = mid - 1;
        }
        nxt[i] = first[r[i]], first[r[i]] = i;
    }
    for(RG int i = 1, g = 0; i <= n + 1; ++i){
        f[i] = g + c[i];
        for(RG int j = first[i]; j != -1; j = nxt[j]) g += w[j];
    }
    RG int ans = f[n + 1];
    for(RG int p = 1; p <= k; ++p){
        Build(1, 1, n);
        for(RG int i = 1; i <= n + 1; ++i){
            f[i] = Query(1, 1, n, 1, i - 1) + c[i];
            for(RG int j = first[i]; j != -1; j = nxt[j])
                if(l[j] > 1) Modify(1, 1, n, 1, l[j] - 1, w[j]);
        }
        ans = min(ans, f[n + 1]);
    }
    printf("%d\n", ans);
    return 0;
}