10239 - The Book-shelver's Problem(dp，卡精度)

d[ i ][ j ]代表当前已决策前i本书，并且当前shelves已经放置j本书，剩下最优决策的最小值；

背包式转移即可；

#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;

const int inf = 1e9;
const int maxn = 1110;
int d[maxn][maxn],W,maxh[maxn][maxn],sumw[maxn];
double sum(int x,int y){
  return sumw[y] - sumw[x-1];
}
struct node{
  int h,w;
  node(int h=0,int w=0):h(h),w(w){}
}book[maxn];
int n;
bool vis[maxn][maxn];
int dp(int i,int j){
  if(vis[i][j]) return d[i][j];
  vis[i][j] = true;
  if(i==n){
     return d[i][j] = maxh[i-j+1][i];
  }
  d[i][j]=dp(i+1,1)+maxh[i-j+1][i];
  if(sum(i-j+1,i+1)<=W){
      d[i][j]=min(d[i][j],dp(i+1,j+1));
  }
  return d[i][j];
}
void scand(int& x){
  int c,d;
  scanf("%d.%d",&c,&d);
  x = c*10000+d;
}
int main()
{
    while(scanf("%d",&n)==1){
      scand(W);
      if(!n) break;
      for(int i=1;i<=n;i++){
        scand(book[i].h);
        scand(book[i].w);
      }
      for(int i=0;i<=n;i++)
        for(int j=0;j<=n;j++)
             vis[i][j] = false;
      for(int i=1;i<=n;i++){
        maxh[i][i]=book[i].h;
        for(int j=i+1;j<=n;j++)
          maxh[i][j]=max(book[j].h,maxh[i][j-1]);
      }
      sumw[0] = 0;
      for(int i=1;i<=n;i++){
        sumw[i]=sumw[i-1]+book[i].w;
      }
      printf("%.4lf\n",dp(1,1)*1.0/10000);
    }
    return 0;
}
/*
5 30.0000
30.0000 20.0000
20.0000 30.0000
25.0000 30.0000
30.0000 25.0000
10.0000 5.0000
*/