BZOJ5311 贞鱼（动态规划+wqs二分+决策单调性）

　　大胆猜想答案随k变化是凸函数，且有决策单调性即可。去粘了份fread快读板子才过。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 4010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=;char c=getchar();
    while (c<''||c>'') c=getchar();
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x;
}
struct FastIO {
    static const int S = 1e7;
    int wpos;
    char wbuf[S];
    FastIO() : wpos() {}
    inline int xchar() {
        static char buf[S];
        static int len = , pos = ;
        if (pos == len)
            pos = , len = fread(buf, , S, stdin);
        if (pos == len) exit();
        return buf[pos++];
    }
    inline int xuint() {
        int c = xchar(), x = ;
        while (c <= ) c = xchar();
        for (; '' <= c && c <= ''; c = xchar()) x = x *  + c - '';
        return x;
    }
    inline int xint()
    {
        int s = , c = xchar(), x = ;
        while (c <= ) c = xchar();
        if (c == '-') s = -, c = xchar();
        for (; '' <= c && c <= ''; c = xchar()) x = x *  + c - '';
        return x * s;
    }
    inline void xstring(char *s)
    {
        int c = xchar();
        while (c <= ) c = xchar();
        for (; c > ; c = xchar()) * s++ = c;
        *s = ;
    }
    inline void wchar(int x)
    {
        if (wpos == S) fwrite(wbuf, , S, stdout), wpos = ;
        wbuf[wpos++] = x;
    }
    inline void wint(ll x)
    {
        if (x < ) wchar('-'), x = -x;
        char s[];
        int n = ;
        while (x || !n) s[n++] = '' + x % , x /= ;
        while (n--) wchar(s[n]);
        wchar('\n');
    }
    inline void wstring(const char *s)
    {
        while (*s) wchar(*s++);
    }
    ~FastIO()
    {
        if (wpos) fwrite(wbuf, , wpos, stdout), wpos = ;
    }
} io;
int n,m,a[N][N],g[N],stk[N],L[N],R[N],top;
ll f[N];
int calc(int i,int j){return a[j][j]-a[i-][j]-a[j][i-]+a[i-][i-]>>;}
bool isbetter(int x,int y,int i){return f[x]+calc(x+,i)<f[y]+calc(y+,i)||f[x]+calc(x+,i)==f[y]+calc(y+,i)&&g[x]<g[y];}
pair<int,ll> check(int cost)
{
    memset(f,,sizeof(f));f[]=;
    stk[top=]=,L[]=,R[]=n;
    for (int i=;i<=n;i++)
    {
        int l=,r=top,x;
        while (l<=r)
        {
            int mid=l+r>>;
            if (R[mid]>=i) x=stk[mid],r=mid-;
            else l=mid+;
        }
        f[i]=f[x]+calc(x+,i)+cost;
        g[i]=g[x]+;
        while (L[top]>i&&isbetter(i,stk[top],L[top])) top--;
        l=i+,r=R[top],x=R[top]+;
        while (l<=r)
        {
            int mid=l+r>>;
            if (isbetter(i,stk[top],mid)) x=mid,r=mid-;
            else l=mid+;
        }
        R[top]=x-;if (x<=n) stk[++top]=i,L[top]=x,R[top]=n;
    }
    return make_pair(g[n],f[n]);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5311.in","r",stdin);
    freopen("bzoj5311.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=;i<=n;i++)
        for (int j=;j<=n;j++)
        a[i][j]=a[i-][j]+a[i][j-]+io.xint()-a[i-][j-];
    int l=-,r=-l,ans;
    while (l<=r)
    {
        int mid=l+r>>;
        if (check(mid).first<=m) ans=mid,r=mid-;
        else l=mid+;
    }
    cout<<check(ans).second-1ll*m*ans;
    return ;
}