【BZOJ1713】[Usaco2007 China]The Bovine Accordion and Banjo Orchestra 音乐会
Description
Input
第1行输入N，之后N行输入Ai，之后N行输入Bi．
Output
输出最大收益．
Sample Input
3
1
1
5
5
1
1
INPUT DETAILS:
There are 6 cows: 3 accordionists and 3 banjoists. The accordionists have
talent levels (1, 1, 5), and the banjoists have talent levels (5, 1, 1).
1
1
5
5
1
1
INPUT DETAILS:
There are 6 cows: 3 accordionists and 3 banjoists. The accordionists have
talent levels (1, 1, 5), and the banjoists have talent levels (5, 1, 1).
Sample Output
17
HINT
手风琴手3和班卓琴手1搭配，创造收益25美元．手风琴手1和手风琴手2喝酒用了4美元．同样班卓琴手2和班卓琴手3用了4美元．最后收益为25 -4-4=17美元．
题解：二维斜率优化，代码不忍直视，注意所有h,t都要用数组来保存，q必须要用二维的
#include <cstdio>
#include <iostream>
#include <cstring>
#define ya(k) (sb[k]*sb[k]-f[i-1][k])
#define xa(k) (sb[k])
#define yb(k) (sa[k]*sa[k]-f[k][j-1])
#define xb(k) (sa[k])
using namespace std;
typedef long long ll;
int n;
ll a[1010],b[1010],qa[1010][1010],qb[1010][1010],sa[1010],sb[1010],ha[1010],hb[1010],ta[1010],tb[1010];
ll f[1010][1010],ans;
int main()
{scanf("%d",&n);int i,j;for(i=1;i<=n;i++) scanf("%lld",&a[i]),sa[i]=sa[i-1]+a[i];for(i=1;i<=n;i++) scanf("%lld",&b[i]),sb[i]=sb[i-1]+b[i];for(i=1;i<=n;i++) ha[i]=hb[i]=1,f[0][i]=-sb[i]*sb[i],f[i][0]=-sa[i]*sa[i];ans=-1,ans<<=60;for(i=1;i<=n;i++){for(j=1;j<=n;j++){f[i][j]=-sa[i-1]*sa[i-1]-sb[j-1]*sb[j-1]+a[i]*b[j];while(ha[i]<ta[i]&&(ya(qa[ha[i]+1][i])-ya(qa[ha[i]][i]))<=(xa(qa[ha[i]+1][i])-xa(qa[ha[i]][i]))*2*sb[j-1]) ha[i]++;if(i>1) f[i][j]=max(f[i][j],f[i-1][qa[ha[i]][i]]+a[i]*b[j]-(sb[j-1]-sb[qa[ha[i]][i]])*(sb[j-1]-sb[qa[ha[i]][i]]));while(ha[i]<ta[i]&&(ya(qa[ta[i]][i])-ya(qa[ta[i]-1][i]))*(xa(j)-xa(qa[ta[i]][i]))>=(ya(j)-ya(qa[ta[i]][i]))*(xa(qa[ta[i]][i])-xa(qa[ta[i]-1][i]))) ta[i]--;qa[++ta[i]][i]=j;while(hb[j]<tb[j]&&(yb(qb[hb[j]+1][j])-yb(qb[hb[j]][j]))<=(xb(qb[hb[j]+1][j])-xb(qb[hb[j]][j]))*2*sa[i-1]) hb[j]++;if(j>1) f[i][j]=max(f[i][j],f[qb[hb[j]][j]][j-1]+a[i]*b[j]-(sa[i-1]-sa[qb[hb[j]][j]])*(sa[i-1]-sa[qb[hb[j]][j]]));while(hb[j]<tb[j]&&(yb(qb[tb[j]][j])-yb(qb[tb[j]-1][j]))*(xb(i)-xb(qb[tb[j]][j]))>=(yb(i)-yb(qb[tb[j]][j]))*(xb(qb[tb[j]][j])-xb(qb[tb[j]-1][j]))) tb[j]--;qb[++tb[j]][j]=i;ans=max(ans,f[i][j]-(sa[n]-sa[i])*(sa[n]-sa[i])-(sb[n]-sb[j])*(sb[n]-sb[j]));}}printf("%lld",ans);return 0;
}
