斜率优化与单调队列DP详解-优快云博客

DP优化之单调队列专题

和单调队列有很多交集。。。

斜率优化理论不说多。一个式子至于变量i有关就极为belong[i];

就是DP方程满足了f[i]=max{belong1[j]+belong1[i]*belong2[j]+belong2[i]}

于是我们就说可以得到斜率belong1[i].以点(belong2[j],belong1[j])组成的凸壳。

求max就是保留上凸，min保留下凸。

---------------------------------------------------------------------

http://acm.hdu.edu.cn/showproblem.php?pid=3045

【HDU3045】【Picnic Cows】【斜率优化】

题目给定数列A，要求排序后，将这些数字分成若干组，每组至少m个数字。

f[i]=min{f[j]+(sum[i]-sum[j]-(i-j+1)*sum[j+1])}

f[i]=min{f[j]-sum[j]+(j-1)*sum[j+1]-i*sum[j+1]+sum[i]}

第一部分只与j有关，第三部分只与i有关，第二部分既与i有关又与i有关。

把-i当做斜率。（sum[j+1],f[j]-sum[j]+(j-1])*sum[j+1])插入凸壳中。因为是求min

所以保留下凸即可。

http://poj.org/problem?id=3709

【pku3709】【K-Anonymous Sequence】【斜率优化】

跟上面那题一样！

http://poj.org/problem?id=2018

【pku2018】【Best Cow Fences】【斜率优化】

题目给定一个一对点(i,f[i])求max{(f[i]-f[j])/(i-j)}

这时候其实是求一个最大斜率，保存一个下凸就可以了。

http://61.187.179.132:8080/JudgeOnline/showproblem?problem_id=1010

【HYOJ1010】【玩具装箱toy】【斜率优化】

题目给定一个序列C。一个长度L

把序列分成若干组使得所有组(sumC+numC-1-L)^2的总和最小。

numC表示该组中元素数，sumC表示该组中C的和。

f[i]=min{f[j]+sqr(sum[i]-sum[j]+(i-j-1)-L)}

f[i]=min{f[j]+sqr(sum[j]+j+1+L)-2*(sum[i]+i)*(sum[j]+j+1+L)+sqr(sum[i]+i)}

第一部分只与j有关，第三部分至于i有关，第二部分与i和j都有关。

以(2*(sum[i]+i))为斜率，以（sum[j]+j+1+L,f[j]+sqr(sum[j]+j+1+L))插入凸壳。

求最小值维护下凸即可。

http://61.187.179.132:8080/JudgeOnline/showproblem?problem_id=1911

【HYOJ1911】【RQNOJ533】【APIO2010】【特别行动部队】【斜率优化】

因为是2次函数，采用相同的方法可以分离系数。分成三部分。

这题很棒！！！

首先n^4的方法是显然的！

f[i][j]=max{f[i'][j']+a[i]*b[j]}

在稍加分析一下会发现

f[i][j]=max{max{f[i-1][j']+a[i]*b[j]},max{f[i'][j-1]}}

再采用斜率优化得到了n^2的方法。于是AC之~

HYOJ1010_玩具装箱toy

#include < cstdio >
#define N 50010
#define LL long long

using namespace std;

LL que[N][ 2 ],sum[N],A[N],B[N],f[N],data[N];
LL n,last,L,tail;
LL sqr(LL x) { return x * x;}
LL cross(LL a[ 2 ],LL b[ 2 ],LL c[ 2 ])
{
return (b[ 0 ] - a[ 0 ]) * (c[ 1 ] - a[ 1 ]) - (b[ 1 ] - a[ 1 ]) * (c[ 0 ] - a[ 0 ]);
}
void hulladd(LL x,LL y)
{
LL node[ 2 ];
node[ 0 ] = x;node[ 1 ] = y;
while (tail >= 2 && cross(node,que[tail],que[tail - 1 ]) >= 0 ) -- tail;
++ tail;que[tail][ 0 ] = node[ 0 ];que[tail][ 1 ] = node[ 1 ];
last <?= tail;
}
LL query(LL x)
{
while (last < tail && x * que[last][ 0 ] + que[last][ 1 ] > x * que[last + 1 ][ 0 ] + que[last + 1 ][ 1 ])
++ last;
return que[last][ 0 ] * x + que[last][ 1 ];
}
int main()
{
freopen( " HYOJ1010.in " , " r " ,stdin);
freopen( " HYOJ1010.out " , " w " ,stdout);
scanf( " %d%d " , & n, & L);last = 1 ;
for (LL i = 1 ;i <= n; ++ i)
{
scanf( " %d " , & data[i]);
sum[i] = sum[i - 1 ] + data[i];
}
for (LL i = 0 ;i <= n; ++ i)
A[i] = sum[i] + i,B[i] = sum[i] + i + 1 + L;
for (LL i = 1 ;i <= n; ++ i)
{
f[i] = sqr(sum[i] + i - L - 1 );
if (i > 1 )
{
hulladd(B[i - 1 ],sqr(B[i - 1 ]) + f[i - 1 ]);
f[i] <?= query( - 2 * A[i]) + sqr(A[i]);
}
}
// for (LL i=1;i<=n;++i) printf("%d ** %d\n",i,f[i]);
printf( " %I64d\n " ,f[n]);
}

pku2018_Best Cow Fences

#include < iostream >
#include < cstdio >
using namespace std;
#define N 100010

int que[N][ 2 ],a[N],sum[N],f[N];
int n,m,temp,tail,ans;
int cross( int a[ 2 ], int b[ 2 ], int c[ 2 ])
{
return (b[ 0 ] - a[ 0 ]) * (c[ 1 ] - a[ 1 ]) - (b[ 1 ] - a[ 1 ]) * (c[ 0 ] - a[ 0 ]);
}
void hulladd( int x, int y)
{
int node[ 2 ];
node[ 0 ] = x;node[ 1 ] = y;
while (tail >= 2 && cross(node,que[tail],que[tail - 1 ]) >= 0 ) -- tail;
++ tail;que[tail][ 0 ] = x;que[tail][ 1 ] = y;
}
int query( int x, int y)
{
int tt = tail,node[ 2 ];
node[ 0 ] = x;node[ 1 ] = y;
while (tt >= 2 && cross(node,que[tt],que[tt - 1 ]) >= 0 ) -- tt;
return 1000 * (y - que[tt][ 1 ]) / (x - que[tt][ 0 ]);
}
int main()
{
freopen( " pku2018.in " , " r " ,stdin);
freopen( " pku2018.out " , " w " ,stdout);
scanf( " %d%d " , & n, & m);
for ( int i = 1 ;i <= n; ++ i) scanf( " %d " , & a[i]);
for ( int i = 1 ;i <= n; ++ i) sum[i] = sum[i - 1 ] + a[i];
for ( int i = m;i <= n; ++ i)
{
f[i] = 1000 * sum[i] / i;
// printf("%d %d \n",i,f[i]);

if (i > m)
{
hulladd(i - m,sum[i - m]);
temp = query(i,sum[i]);
if (temp > f[i]) f[i] = temp;
}
// printf("%d %d \n",i,f[i]);
}
for ( int i = m;i <= n; ++ i)
{
// printf("%d %d\n",i,f[i]);
if (f[i] > ans) ans = f[i];
}
printf( " %d " ,ans);
return 0 ;
}

HDU3045_Picnic Cows

#include < iostream >
#include < cstdio >
#define N 400010
#define LL long long
using namespace std;

LL tail,a[N],que[N][ 2 ],sum[N],n,m,last,f[N],i,j,t;

LL cross(LL a[ 2 ],LL b[ 2 ],LL c[ 2 ])
{
return (b[ 0 ] - a[ 0 ]) * (c[ 1 ] - a[ 1 ]) - (b[ 1 ] - a[ 1 ]) * (c[ 0 ] - a[ 0 ]);
}
void hulladd(LL x,LL y)
{
LL node[ 2 ];
node[ 0 ] = x;node[ 1 ] = y;
while (tail >= 1 && x == que[tail][ 0 ] && y < que[tail][ 1 ]) -- tail;
if (tail > 0 && que[tail][ 0 ] == x && y >= que[tail][ 1 ]) return ;
while (tail >= 2 && cross(node,que[tail],que[tail - 1 ]) >= 0 ) -- tail;
++ tail;que[tail][ 0 ] = node[ 0 ];que[tail][ 1 ] = node[ 1 ];
last <?= tail;
}
LL query(LL x)
{
last = tail;
while (last > 1 && que[last][ 0 ] * x + que[last][ 1 ] > que[last - 1 ][ 0 ] * x + que[last - 1 ][ 1 ])
-- last;
return que[last][ 0 ] * x + que[last][ 1 ];
}
void qsort(LL l,LL r)
{
LL mid = a[(l + r) / 2 ],i = l,j = r;
while (i < j)
{
while (a[i] < mid) ++ i;
while (a[j] > mid) -- j;
if (i <= j)
{
t = a[i];a[i] = a[j];a[j] = t;
++ i; -- j;
}
}
if (i < r) qsort(i,r);
if (l < j) qsort(l,j);
}
int main()
{
freopen( " HDU3045.in " , " r " ,stdin);
freopen( " HDU3045.out " , " w " ,stdout);

while (scanf( " %d%d " , & n, & m) != EOF)
{
for (LL i = 0 ;i < n; ++ i) scanf( " %I64d " , & a[i]);
qsort( 0 ,n - 1 );
for (LL i = n;i >= 1 ; -- i) a[i] = a[i - 1 ];a[ 0 ] = 0 ;
for (LL i = 0 ;i <= n; ++ i) sum[i] = 0 ;
for (LL i = 1 ;i <= n; ++ i) sum[i] = sum[i - 1 ] + a[i];
tail = 0 ;last = 1 ;
if (m == 0 ) {printf( " 0\n " ); continue ;}
for (LL i = m;i <= n; ++ i)
{
f[i] = sum[i] - i * a[ 1 ];
if (i == n)
{
i = n;
}
if (i - m >= m)
{
hulladd(a[i - m + 1 ],f[i - m] - sum[i - m] + (i - m) * a[i - m + 1 ]);
f[i] <?= query( - i) + sum[i];
}
}
printf( " %I64d\n " ,f[n]);
}
return 0 ;
}

pku3709_K-Anonymous Sequence

#include < iostream >
#include < cstdio >
#define N 500010
#define LL long long
using namespace std;

LL tail,a[N],que[N][ 2 ],sum[N],n,m,last,f[N],i,j,t,T,temp;

LL cross(LL a[ 2 ],LL b[ 2 ],LL c[ 2 ])
{
return (b[ 0 ] - a[ 0 ]) * (c[ 1 ] - a[ 1 ]) - (b[ 1 ] - a[ 1 ]) * (c[ 0 ] - a[ 0 ]);
}
void hulladd(LL x,LL y)
{
LL node[ 2 ];
node[ 0 ] = x;node[ 1 ] = y;
while (tail >= 1 && x == que[tail][ 0 ] && y < que[tail][ 1 ]) -- tail;
if (tail > 0 && que[tail][ 0 ] == x && y >= que[tail][ 1 ]) return ;
while (tail >= 2 && cross(node,que[tail],que[tail - 1 ]) >= 0 ) -- tail;
++ tail;que[tail][ 0 ] = node[ 0 ];que[tail][ 1 ] = node[ 1 ];
}
LL query(LL x)
{
last = tail;
while (last > 1 && que[last][ 0 ] * x + que[last][ 1 ] > que[last - 1 ][ 0 ] * x + que[last - 1 ][ 1 ])
-- last;
return que[last][ 0 ] * x + que[last][ 1 ];
}
void qsort(LL l,LL r)
{
LL mid = a[(l + r) / 2 ],i = l,j = r;
while (i < j)
{
while (a[i] < mid) ++ i;
while (a[j] > mid) -- j;
if (i <= j)
{
t = a[i];a[i] = a[j];a[j] = t;
++ i; -- j;
}
}
if (i < r) qsort(i,r);
if (l < j) qsort(l,j);
}
int main()
{
freopen( " pku3709.in " , " r " ,stdin);
freopen( " pku3709.out " , " w " ,stdout);
scanf( " %d " , & T);

while (scanf( " %d%d " , & n, & m) != EOF)
{
for (LL i = 0 ;i < n; ++ i) scanf( " %I64d " , & a[i]);
qsort( 0 ,n - 1 );
for (LL i = n;i >= 1 ; -- i) a[i] = a[i - 1 ];a[ 0 ] = 0 ;
for (LL i = 0 ;i <= n; ++ i) sum[i] = 0 ;
for (LL i = 1 ;i <= n; ++ i) sum[i] = sum[i - 1 ] + a[i];
tail = 0 ;last = 1 ;
if (m == 0 ) {printf( " 0\n " ); continue ;}
for (LL i = m;i <= n; ++ i)
{
f[i] = sum[i] - i * a[ 1 ];
if (i == n)
{
i = n;
}
if (i - m >= m)
{
hulladd(a[i - m + 1 ],f[i - m] - sum[i - m] + (i - m) * a[i - m + 1 ]);
temp = query( - i) + sum[i];
if (temp < f[i]) f[i] = temp;
}
}
printf( " %I64d\n " ,f[n]);
}

return 0 ;
}

HYOJ1911_commando

#include < iostream >
#include < cstdio >
#define N 1001000
#define LL long long
using namespace std;

LL tail,last;
LL que[N][ 2 ],s[N],f[N],ans,data[N],n,a,b,c;
LL sqr(LL x) { return x * x;}
LL cross(LL a[ 2 ],LL b[ 2 ],LL c[ 2 ])
{
return (b[ 0 ] - a[ 0 ]) * (c[ 1 ] - a[ 1 ]) - (b[ 1 ] - a[ 1 ]) * (c[ 0 ] - a[ 0 ]);
}
void hulladd(LL x,LL y)
{
LL node[ 2 ];
// printf("***%I64d %I64d\n",x,y);
node[ 0 ] = x;node[ 1 ] = y;
while (tail >= 2 && cross(node,que[tail],que[tail - 1 ]) <= 0 ) -- tail;
++ tail;que[tail][ 0 ] = node[ 0 ];que[tail][ 1 ] = node[ 1 ];
last <?= tail;
}
LL query(LL x)
{
// ++last;last<?=tail;
// last=tail;
while (last < tail && que[last][ 0 ] * x + que[last][ 1 ] < que[last + 1 ][ 0 ] * x + que[last + 1 ][ 1 ]) ++ last;
while (last > 1 && que[last][ 0 ] * x + que[last][ 1 ] < que[last - 1 ][ 0 ] * x + que[last - 1 ][ 1 ]) -- last;
// printf("%I64d %I64d %I64d\n",x,last,tail);
// if (last+1<tail) {printf("--------");}
return que[last][ 0 ] * x + que[last][ 1 ];

}
int main()
{
freopen( " rq533.in " , " r " ,stdin);
freopen( " rq533.out " , " w " ,stdout);
scanf( " %I64d " , & n);last = 1 ;tail = 0 ;
scanf( " %I64d%I64d%I64d " , & a, & b, & c);

for (LL i = 1 ;i <= n; ++ i) scanf( " %I64d " , & data[i]);
for (LL i = 1 ;i <= n; ++ i) s[i] = s[i - 1 ] + data[i];
for (LL i = 1 ;i <= n; ++ i)
{

f[i] = a * sqr(s[i]) + b * s[i] + c;
// printf("%I64d\n",f[i]);
if (i > 1 )
{
hulladd(s[i - 1 ],a * sqr(s[i - 1 ]) + f[i - 1 ]);
f[i] >?= query( - ( 2 * a * s[i] + b)) + c + a * sqr(s[i]) + b * s[i];
}
// printf("%I64d\n",f[i]);
}
// for (int i=1;i<=n;++i) printf("%d\n",f[i]);
printf( " %I64d\n " ,f[n]);
return 0 ;
}

HYOJ1713_The Bovine Accordion and Banjo Orchestra

#include < iostream >
#include < cstdio >
#define N 1001
#define LL long long
using namespace std;

LL ans,last[N],que[N][N][ 2 ],a[N],b[N],sa[N],sb[N],n,tail[N],f[N][N];
LL sqr(LL x) { return x * x;}
LL cross(LL a[ 2 ],LL b[ 2 ],LL c[ 2 ])
{
return (b[ 0 ] - a[ 0 ]) * (c[ 1 ] - a[ 1 ]) - (b[ 1 ] - a[ 1 ]) * (c[ 0 ] - a[ 0 ]);
}
void hulladd(LL id,LL x,LL y)
{
LL node[ 2 ];
node[ 0 ] = x;node[ 1 ] = y;
while (tail[id] >= 2 && cross(node,que[id][tail[id]],que[id][tail[id] - 1 ]) <= 0 ) -- tail[id];
++ tail[id];
que[id][tail[id]][ 0 ] = x;que[id][tail[id]][ 1 ] = y;
last[id] <?= tail[id];
}
LL query(LL id,LL t)
{
LL temp1,temp2;
while (last[id] > 1 && que[id][last[id]][ 0 ] * t + que[id][last[id]][ 1 ] < que[id][last[id] - 1 ][ 0 ] * t + que[id][last[id] - 1 ][ 1 ])
-- last[id];
while (last[id] < tail[id] && (temp1 = que[id][last[id]][ 0 ] * t + que[id][last[id]][ 1 ]) < (temp2 = que[id][last[id] + 1 ][ 0 ] * t + que[id][last[id] + 1 ][ 1 ]))
++ last[id];
return que[id][last[id]][ 0 ] * t + que[id][last[id]][ 1 ];
}
int main()
{
freopen( " HYOJ1713.in " , " r " ,stdin);
freopen( " HYOJ1713.out " , " w " ,stdout);
scanf( " %d " , & n);
ans =- 99999 ;
for (LL i = 1 ;i <= n; ++ i) scanf( " %d " , & a[i]),sa[i] = sa[i - 1 ] + a[i];
for (LL i = 1 ;i <= n; ++ i) scanf( " %d " , & b[i]),sb[i] = sb[i - 1 ] + b[i];
for (LL i = 1 ;i <= n; ++ i) tail[i] = 0 ,last[i] = 1 ;
for (LL i = 1 ;i <= n; ++ i)
{
tail[ 0 ] = 0 ;last[ 0 ] = 1 ;
for (LL j = 1 ;j <= n; ++ j)
{
f[i][j] = a[i] * b[j] - sqr(sa[i - 1 ]) - sqr(sb[j - 1 ]);
if (i > 1 )
{
if (j > 1 )
{
if (i == 4 && j == 3 )
{
i = 4 ;
}
hulladd( 0 ,sb[j - 1 ],f[i - 1 ][j - 1 ] - sqr(sb[j - 1 ]));
f[i][j] >?= query( 0 , 2 * sb[j - 1 ]) + a[i] * b[j] - sqr(sb[j - 1 ]);
hulladd(j - 1 ,sa[i - 1 ],f[i - 1 ][j - 1 ] - sqr(sa[i - 1 ]));
f[i][j] >?= query(j - 1 , 2 * sa[i - 1 ]) + a[i] * b[j] - sqr(sa[i - 1 ]);
}
}
ans >?= f[i][j] - sqr(sa[n] - sa[i]) - sqr(sb[n] - sb[j]);
}
}
// for (LL i=1;i<=n;++i)
// for (LL j=1;j<=n;++j)
// prLLf("%d %d %d\n",i,j,f[i][j]);
printf( " %d\n " ,ans);
return 0 ;
}

转载于:https://www.cnblogs.com/LitIce/archive/2010/10/12/1848651.html