hdu5550 Game Rooms

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 130    Accepted Submission(s): 39

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 Ti 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.

 

这题是一道dp题，思路很难想，看了被人的博客后才做了出来，具体看代码中的解释。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 0x7fffffff
#define maxn 4050
ll a[maxn],b[maxn];
ll sum[maxn][2];//sum[i][u]表示前i层楼，性别为u的总人数
ll dsum[maxn][2];//dsum[i][u]表示前i层楼，性别为u者距离0层的距离之和
ll dp[maxn][2];//dp[i][u]表示第i层为u属性，第i+1层为另一属性，前i层不同性别到达自己的最近属性的寝室的最近距离和

ll goup(int l,int r,int sex){   //表示[l+1,r]区间sex性别要去r+1的总距离
    return (sum[r][sex]-sum[l][sex])*(r+1)-(dsum[r][sex]-dsum[l][sex]);
}

ll godown(int l,int r,int sex){  //表示[l+1,r]区间sex性别要去l的总距离
    return dsum[r][sex]-dsum[l][sex]-(sum[r][sex]-sum[l][sex])*l;
}

ll cnt(int l,int r,int sex){   //在[l,r]都是sex属性，且l-1与r+1都为非sex属性的条件下。 [l,r]这些楼层非sex属性的人，去自己属性寝室的最小距离。
    int mid=(l+r)>>1;
    return godown(l-1,mid,sex)+goup(mid,r,sex);
}
int main()
{
    int n,m,i,j,T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        sum[0][0]=sum[0][1]=0;
        dsum[0][0]=dsum[0][1]=0;
        for(i=1;i<=n;i++){
            scanf("%lld%lld",&a[i],&b[i]);
            sum[i][0]=sum[i-1][0]+a[i];
            sum[i][1]=sum[i-1][1]+b[i];
            dsum[i][0]=dsum[i-1][0]+a[i]*i;
            dsum[i][1]=dsum[i-1][1]+b[i]*i;
        }
        memset(dp,0,sizeof(dp));
        ll ans=1e18;
        for(i=1;i<n;i++){
            dp[i][0]=goup(0,i,1);  //这里先假设前i层都是性别0，i+1层是性别1所要的总距离
            dp[i][1]=goup(0,i,0);
            for(j=1;j<i;j++){
                dp[i][0]=min(dp[i][0],dp[j][1]+cnt(j+1,i,1)   ); //依次使得前j层是1，使得第j,i+1都是1，这样就好状态转移了
                dp[i][1]=min(dp[i][1],dp[j][0]+cnt(j+1,i,0)   );
            }
            ans=min(ans,dp[i][0]+godown(i,n,0)  ); //这里每一层都要更新一下ans，而不能最后才更新，因为最后才更新的话就不能使得后面几层都相同了
            ans=min(ans,dp[i][1]+godown(i,n,1)  );
        }
        cas++;
        printf("Case #%d: %lld
",cas,ans);
    }
    return 0;
}