POJ1505:Copying Books(区间DP)
Description
Once upon a time, there was a theater ensemble that wanted to play famous Antique Tragedies. The scripts of these plays were divided into many books and actors needed more copies of them, of course. So they hired many scribers to make copies of these books. Imagine you have m books (numbered 1, 2 ... m) that may have different number of pages (p1, p2 ... pm) and you want to make one copy of each of them. Your task is to divide these books among k scribes, k <= m. Each book can be assigned to a single scriber only, and every scriber must get a continuous sequence of books. That means, there exists an increasing succession of numbers 0 = b0 < b1 < b2, ... < b
k-1 <= bk = m such that i-th scriber gets a sequence of books with numbers between bi-1+1 and bi. The time needed to make a copy of all the books is determined by the scriber who was assigned the most work. Therefore, our goal is to minimize the maximum number of pages assigned to a single scriber. Your task is to find the optimal assignment.
Input
Output
If there is more than one solution, print the one that minimizes the work assigned to the first scriber, then to the second scriber etc. But each scriber must be assigned at least one book.
Sample Input
2
9 3
100 200 300 400 500 600 700 800 900
5 4
100 100 100 100 100
Sample Output
100 200 300 400 500 / 600 700 / 800 900
100 / 100 / 100 / 100 100
题意:有一个序列的书分给n个人,区间最大的最小,有多重分法,就让和大的区间尽量靠后
思路;明显的区间DP
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; int dp[505][505],sum[505],a[505]; int main()
{
int t,n,m,i,j,x,v,cas;
scanf("%d",&cas);
while(cas--)
{
scanf("%d%d",&n,&m);
memset(sum,0,sizeof(sum));
memset(dp,-1,sizeof(dp));
for(i = 1; i<=n; i++)
{
scanf("%d",&x);
sum[i] = sum[i-1]+x;
}
dp[0][0] = 0;
for(i = 1; i<=n; i++)
{
for(j = 1; j<=i && j<=m; j++)
{
if(j == 1)
dp[i][j] = sum[i];
else
{
for(v = j-1; v<=i-1; v++)//
{
int t = max(dp[v][j-1],sum[i]-sum[v]);
if(dp[i][j] == -1 || t<dp[i][j])
dp[i][j] = t;
}
}
}
}
j = m-1;
x = 0;
for(i = n; i>=1; i--)
{
x+=sum[i]-sum[i-1];
if(x>dp[n][m] || i<=j)
{
a[j--] = i+1;
x = sum[i]-sum[i-1];
}
}
int cnt = 1;
for(i = 1; i<=n; i++)
{
if(i>1)
printf(" ");
if(cnt<m && a[cnt]==i)
{
printf("/ ");
cnt++;
}
printf("%d",sum[i]-sum[i-1]);
}
printf("\n");
} return 0;
}
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