## 【POJ 1821】Fence

### 【题解大意】

1.当j变大时，b把小于j-L的决策出队；

2.有新的决策入队时，在队尾检查f[i-1,k]的单调性，把无用决策从队尾直接出队，然后把新决策入队。

### 【code】

```#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define File "fence"
inline void file(){
freopen(File".in","r",stdin);
freopen(File".out","w",stdout);
}
int x = 0,f = 1; char ch = getchar();
while(ch < '0' || ch > '9'){if(ch == '-')f = -1; ch = getchar();}
while(ch >= '0' && ch <= '9'){x = (x<<1) + (x<<3) + ch-'0'; ch = getchar();}
return x*f;
}
const int mxn = 16000 + 10;
const int mxk = 105;
int n,k;

struct P{
int l,v,s;
}p[mxk];
inline bool cmp(P t1,P t2){
return t1.s < t2.s;
}

int q[mxn];
int f[mxk][mxn];
inline int Calc(int i,int j){
return f[i-1][j] - p[i].v*j;
}

int main(){
file();
for(int i = 1;i <= k; ++i)

sort(p+1,p+k+1,cmp);

int l(1),r(0);
for(int i = 1;i <= k; ++i){
l = 1,r = 0;
for(int j = max(p[i].s-p[i].l,0);j <= p[i].s-1; ++j){
while(l <= r && Calc(i,q[r]) <= Calc(i,j)) --r;
q[++r] = j;//新的决策从队尾入队
}
for(int j = 1;j <= n; ++j){
f[i][j] = max(f[i-1][j],f[i][j-1]);
if(j >= p[i].s){
while(l <= r && q[l] < j - p[i].l) ++l;
if(l <= r) f[i][j] = max(f[i][j],Calc(i,q[l]) + p[i].v*j);
}
}
}
printf("%d\n",f[k][n]);
return 0;
}```
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