洛谷 P3486 [POI2009]KON-Ticket Inspector
题目描述
Byteasar works as a ticket inspector in a Byteotian National Railways (BNR) express train that connects Byteburg with Bitwise.
The third stage of the BNR reform (The never ending saga of BNR reforms and the Bitwise hub was presented in the problems Railway from the third stage of XIV Polish OI and Station from the third stage of XV Polish OI. Their knowledge, however, is not required at all in order to solve this problem.) has begun. In particular, the salaries system has already been changed.
For example, to encourage Byteasar and other ticket inspectors to efficient work, their salaries now depend on the number of tickets (passengers) they inspect. Byteasar is able to control all the passengers on the train in the time between two successive stations, but he is not eager to waste his energy in doing so. Eventually he decided he would check the tickets exactly 
times per ride.
Before setting out, Byteasar is given a detailed summary from which he knows exactly how many passengers will travel from each station to another. Based on that he would like to choose the moments of control in such a way that the number of passengers checked is maximal. Obviously, Byteasar is not paid extra for checking someone multiple times - that would be pointless, and would only disturb the passengers. Write a programme that will determine for Byteasar when he should check the tickets in order to maximise his revenue.
有n个车站，现在有一辆火车从1到n驶过，给出aij代表从i站上车j站下车的人的个数。列车行驶过程中你有K次检票机会，所有当前在车上的人会被检票，问最多能检多少个不同的人的票
输入输出格式
输入格式：
In the first line of the standard input two positive integers 
and 
(
, 
) are given. These are separated by a single space and denote, respectively, the number of stations en route and the number of controls Byteasar intends to make. The stations are numbered from 
to 
in the order of appearance on the route.
In the next 
lines the summary on passengers is given. The 
-th line contains information on the passengers who enter the train on the station 
- it is a sequence of  an increasing sequence of 
integers from the interval from 
to 
separated by single spaces to the standard output. These numbers should be the numbers of stations upon leaving which Byteasar should control the tickets.
输入输出样例
7 2
2 1 8 2 1 0
3 5 1 0 1
3 1 2 2
3 5 6
3 2
1
2 5
思路：dp
设f[i][j]表示第i次检票在第j车站的最优解。
那么就可以很容易列出状态转移方程：
sum是一个关于上车人数的一维前缀和。
summ是一个关于下车人数的二维前缀和。
f[i][j]=max(f[i][j],f[i-1][l]+sum[j]-sum[l]-tot);(tot=summ[j][j]-summ[l][j])
错因：在计算tot时的式子表示错了。
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; int n,k,ans,pre; int sum[610],an[60]; int step[60][610],f[60][601]; int map[610][610],summ[610][610]; int main(){ scanf("%d%d",&n,&k); for(int i=1;i<=n;i++){ for(int j=1;j<=i;j++) summ[i][j]+=summ[i-1][j]+summ[i][j-1]-summ[i-1][j-1]; for(int j=i+1;j<=n;j++){ scanf("%d",&map[i][j]); sum[i]+=map[i][j]; summ[i][j]+=map[i][j]+summ[i-1][j]+summ[i][j-1]-summ[i-1][j-1]; } sum[i]+=sum[i-1]; } for(int i=1;i<=n;i++) f[1][i]=sum[i]-summ[i][i]-summ[i][0]-summ[0][i]+summ[0][0]; for(int i=2;i<=k;i++) for(int j=i;j<=(n-k+i);j++) for(int l=i-1;l<=j-1;l++){ int tot=summ[j][j]-summ[l][j]; if(f[i][j]<f[i-1][l]+sum[j]-sum[l]-tot){ step[i][j]=l; f[i][j]=f[i-1][l]+sum[j]-sum[l]-tot; } } for(int i=k;i<=n;i++) if(f[k][i]>ans){ ans=f[k][i];pre=i; } an[k]=pre; for(int i=k;i>1;i--){ an[i-1]=step[i][pre]; pre=step[i][pre]; } for(int i=1;i<=k;i++) cout<<an[i]<<" "; } /* 7 2 2 1 8 2 1 0 3 5 1 0 1 3 1 2 2 3 5 6 3 2 1 */
