題意：每層樓有ai個人打羽毛球，有bi個人游泳，每層樓只能建造一個羽毛球場或者游泳館，如果當前樓層沒有羽毛求場，這ai個人就要到最近有羽毛球場的樓層打羽毛球，每個人產生的代價為樓層差，游泳的類似，求最小代價

這個貌似並沒有用到什麼很特殊的技巧可是就是莫名地難想。。不造為什麼。。

首先由於樓層上下的情況都會對當前樓層產生影響，所以不能一層一層地往上考慮。。

那就只能一段一段考慮了。。對一段區間[l,r]，假設他都用於建造羽毛球場，那麼游泳的人由區間中點分開分別到l-1和r+1去游泳，所以當一段區間用於建造羽毛球或者游泳館，我們是可以預處理的。。中點分開用字首差分一下就可以了。。

預處理完用DP成段更新即可。。不過要注意第一段和最後一段的人只能往一個方向移動，所以應該單獨出來討論處理。。

 

 

 

#include<bits/stdc++.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge*j=h[x];j;j=j->next)
#define eps 1e-8
#define mid (x+y>>1)
#define lowbit(x) (x&(-x))
#define ll long long
#define sqr(x) ((x)*(x))
#define mem(a) memset(a,0,sizeof(a))
#define succ(x) (1<<(x))
#define NM 4005
using namespace std;
const ll inf=1e18;
const double pi=acos(-1);
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return f*x;
}











int n,ca;
ll a[NM],b[NM],_a[NM],_b[NM],f[NM],g[NM],ans;


ll c(int i,int j){
    int t=i+j>>1;
    return _a[t]-_a[i-1]-(i-1)*(a[t]-a[i-1])+(a[j]-a[t])*(j+1)-(_a[j]-_a[t]);
}
ll d(int i,int j){
    int t=i+j>>1;
    return _b[t]-_b[i-1]-(i-1)*(b[t]-b[i-1])+(b[j]-b[t])*(j+1)-(_b[j]-_b[t]);
}

int main(){
    //freopen("data.in","r",stdin);
    int _=read();while(_--){
	n=read();
	inc(i,1,n)a[i]=read(),b[i]=read();
	inc(i,1,n)_a[i]=a[i]*i,_b[i]=b[i]*i;
	inc(i,1,n)_a[i]+=_a[i-1],_b[i]+=_b[i-1],a[i]+=a[i-1],b[i]+=b[i-1];
	inc(i,1,n-1)g[i]=(i+1)*b[i]-_b[i],f[i]=(i+1)*a[i]-_a[i];
	g[n]=f[n]=inf;
	inc(i,2,n){
	    inc(j,1,i-1)f[i]=min(f[i],g[j]+c(j+1,i)),g[i]=min(g[i],f[j]+d(j+1,i));
	}
	ans=inf;
	//inc(i,1,n)inc(j,i,n)printf("%d %d:%lld %lld\n",i,j,c(i,j),d(i,j));
	inc(i,1,n-1){
	    ans=min(ans,g[i]+(_a[n]-_a[i])-i*(a[n]-a[i]));
	    ans=min(ans,f[i]+(_b[n]-_b[i])-i*(b[n]-b[i]));
	}
	printf("Case #%d: %lld\n",++ca,ans);
    }
    return 0;
}
 

 

 

 

 

 

Game Rooms

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 989    Accepted Submission(s): 321

 

Problem Description

Your company has just constructed a new skyscraper, but you just noticed a terrible problem: there is only space to put one game room on each floor! The game rooms have not been furnished yet, so you can still decide which ones should be for table tennis and which ones should be for pool. There must be at least one game room of each type in the building.

Luckily, you know who will work where in this building (everyone has picked out offices). You know that there will be T

i table tennis players and Pi pool players on each floor. Our goal is to minimize the sum of distances for each employee to their nearest game room. The distance is the difference in floor numbers: 0 if an employee is on the same floor as a game room of their desired type, 1 if the nearest game room of the desired type is exactly one floor above or below the employee, and so on.

 

 

Input

The first line of the input gives the number of test cases, T(1≤T≤100). T test cases follow. Each test case begins with one line with an integer N(2≤N≤4000), the number of floors in the building. N lines follow, each consists of 2 integers, Ti and Pi(1≤Ti,Pi≤109), the number of table tennis and pool players on the ith floor. The lines are given in increasing order of floor number, starting with floor 1 and going upward.

 

 

Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the minimal sum of distances.

 

 

Sample Input

1 2 10 5 4 3

 

 

Sample Output

Case #1: 9

Hint

In the first case, you can build a table tennis game room on the first floor and a pool game room on the second floor. In this case, the 5 pool players on the first floor will need to go one floor up, and the 4 table tennis players on the second floor will need to go one floor down. So the total distance is 9.

 

 

Source

The 2015 China Collegiate Programming Contest

 

 

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