Uva--10154（动规）

最新推荐文章于 2020-10-29 11:24:19 发布

转载最新推荐文章于 2020-10-29 11:24:19 发布 · 81 阅读

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原文链接：http://www.cnblogs.com/naturepengchen/articles/3891322.html

本文介绍了一个关于构建最大龟塔的算法问题。该问题通过动态规划解决如何在考虑到每个龟的重量和力量的前提下，构建最高的龟塔。文章提供了完整的代码实现，并解释了DP方程。

2014-08-04 23:50:01

Problem F: Weights and Measures

I know, up on top you are seeing great sights,
But down at the bottom, we, too, should have rights.
We turtles can't stand it. Our shells will all crack!
Besides, we need food. We are starving!" groaned Mack.

The Problem

Mack, in an effort to avoid being cracked, has enlisted your advice as to the order in which turtles should be dispatched to form Yertle's throne. Each of the five thousand, six hundred and seven turtles ordered by Yertle has a different weight and strength. Your task is to build the largest stack of turtles possible.

Standard input consists of several lines, each containing a pair of integers separated by one or more space characters, specifying the weight and strength of a turtle. The weight of the turtle is in grams. The strength, also in grams, is the turtle's overall carrying capacity, including its own weight. That is, a turtle weighing 300g with a strength of 1000g could carry 700g of turtles on its back. There are at most 5,607 turtles.

Your output is a single integer indicating the maximum number of turtles that can be stacked without exceeding the strength of any one.

Sample Input

Sample Output

3

思路：关键在看问题的角度，应该从stack的顶上往下考虑，这样递推比较自然，将龟龟的力量从小到大排个序。DP方程：dp[i][j]表示用前 i 个龟龟叠 j 层高的最低总重量。
　　DP方程：dp[i][j] = min(dp[i][j] , dp[i - 1][j] , dp[i - 1][j - 1] + w[i])

 1 /*************************************************************************
 2     > File Name: g.cpp
 3     > Author: Nature
 4     > Mail: 564374850@qq.com
 5     > Created Time: Sun 03 Aug 2014 09:44:36 PM CST
 6 ************************************************************************/
 7 
 8 #include <cstdio>
 9 #include <cstring>
10 #include <cstdlib>
11 #include <cmath>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 const int INF = 1e9;
16 
17 struct node{
18     int w,s;
19 }no[6000];
20 
21 int cnt = 1;
22 int dp[6000][6000];
23 
24 bool cmp(node a,node b){
25     return a.s - a.w < b.s - b.w;
26 }
27 
28 int main(){
29     //freopen("in","r",stdin);
30     while(scanf("%d%d",&no[cnt].w,&no[cnt].s) == 2)
31         ++cnt;
32     sort(no + 1,no + cnt,cmp);
33     //printf("%d\n",cnt);
34     int ansmax = 0;
35     for(int i = 0; i < cnt; ++i)
36         for(int j = 1; j < cnt; ++j)
37             dp[i][j] = INF;
38     for(int i = 1; i < cnt; ++i){
39         for(int j = 1; j <= i; ++j){
40             dp[i][j] = dp[i - 1][j];
41             if(no[i].s - no[i].w - dp[i - 1][j - 1] >= 0)
42                 dp[i][j] = min(dp[i][j],dp[i - 1][j - 1] + no[i].w);
43             if(dp[i][j] < INF){
44                 //printf("dp[%d][%d] : %d\n",i,j,dp[i][j]);
45                 ansmax = max(ansmax,j);
46             }
47         }
48     }
49     printf("%d\n",ansmax);
50 }

转载于:https://www.cnblogs.com/naturepengchen/articles/3891322.html