ACM: 动态规划题 toj 1545

Hurdles of 110m

描述

 
In the year 2008, the 29th
Olympic Games will be held in Beijing. This will signify the
prosperity of China and Beijing Olympics is to be a festival for
people all over the world as well.

Liu Xiang is one of the famous Olympic athletes in China. In
2002 Liu broke Renaldo Nehemiah's 24-year-old world junior record
for the 110m hurdles. At the 2004 Athens Olympics Games, he won the
gold medal in the end. Although he was not considered as a favorite
for the gold, in the final, Liu's technique was nearly perfect as
he barely touched the sixth hurdle and cleared all of the others
cleanly. He powered to a victory of almost three meters. In doing
so, he tied the 11-year-old world record of 12.91 seconds. Liu was
the first Chinese man to win an Olympic gold medal in track and
field. Only 21 years old at the time of his victory, Liu vowed to
defend his title when the Olympics come to Beijing in 2008.



In the 110m hurdle competition, the track was divided into
N parts by the hurdle. In each part, the player has to run
in the same speed; otherwise he may hit the hurdle. In fact, there
are 3 modes to choose in each part for an athlete -- Fast Mode,
Normal Mode and Slow Mode. Fast Mode costs the player T1
time to pass the part. However, he cannot always use this mode in
all parts, because he needs to consume F1 force at the same
time. If he doesn't have enough force, he cannot run in the part at
the Fast Mode. Normal Mode costs the player T2 time for the
part. And at this mode, the player's force will remain unchanged.
Slow Mode costs the player T3 time to pass the part.
Meanwhile, the player will earn F2 force as compensation.
The maximal force of a player is M. If he already has
M force, he cannot earn any more force. At the beginning of
the competition, the player has the maximal force.

The input of this problem is detail data for Liu Xiang. Your
task is to help him to choose proper mode in each part to finish
the competition in the shortest time.

输入

 
Standard input will
contain multiple test cases. The first line of the input is a
single integer T (1 <= T
<= 50) which is the number of test cases. And it
will be followed by T consecutive test cases.

Each test case begins with two positive integers N and
M. And following N lines denote the data for the
N parts. Each line has five positive integers T1 T2 T3 F1
F2. All the integers in this problem are less than or equal to
110.

输出

Results
should be directed to standard output. The output of each test case
should be a single integer in one line, which is the shortest time
that Liu Xiang can finish the competition.

样例输入
2

1 10

1 2 3 10 10

4 10

1 2 3 10 10

1 10 10 10 10

1 1 2 10 10

1 10 10 10 10
样例输出
1

6
提示
For the second sample test case, Liu Xiang should
run with the sequence of Normal Mode, Fast Mode, Slow Mode and Fast
Mode.

题意: 刘翔跑110栏, 经过n个栏,
他每次经过一个栏有三种模式:

         
fast: 消耗force的,  normal: 什么也不消耗, slow: 增加force.
一开始的时候force是最大值m.

        
问最小的时间跑完全程.
解题思路:

                 
1. DP题先找状态:  dp[i][j]: 经过前i个栏剩下j的体力最小的时间.

                 
2. 决策: 

                             
(1). fast:  当j >= a[i].f1 时,
dp[i+1][ j-a[i].f1 ] = min(dp[i+1][ j-a[i].f1 ] ,
dp[i][j]+a[i].t1);

                             
(2). normal: 任何情况都满足 dp[i+1][j] = min(dp[i+1][j] ,
dp[i][j]+a[i].t2);

                             
(3). slow: 任何情况都满足 (但是有两种情况.)

                                          体力恢复没有超过m值:
dp[i+1][ j+a[i].f2 ] = min(dp[i+1][ j+a[i].f2 ] ,
dp[i][j]+a[i].t3);

                                         
体力恢复超过m值: dp[i+1][m] = min(dp[i+1][m] , dp[i][j]+a[i].t3);

                 
3. 最从dp
[k]中找出最小值即可.  (0<= k
<= m);
 
代码:
#include
<cstdio>

#include <iostream>

#include <cstring>

using namespace std;

#define MAX 115

const int INF = (1<<29);

struct
node

{

 int t1, t2, t3;

 int f1, f2;

}a[MAX];
int n,
m;

int dp[MAX][MAX];

 

inline int min(int a,int b)

{

 return a < b ? a : b;

}

int main()

{

 int i, j;

// freopen("input.txt","r",stdin);

 int caseNum;

 scanf("%d",&caseNum);

 while(caseNum--)

 {

  scanf("%d
%d",&n,&m);

  for(i = 0; i <
n; ++i)

   scanf("%d %d
%d %d
%d",&a[i].t1,&a[i].t2,&a[i].t3,&a[i].f1,&a[i].f2);

  for(i = 0; i
<= n; ++i)

   for(j = 0; j
<= m; ++j)

    dp[i][j]
= INF;

  dp[0][m] = 0;

  for(i = 0; i
< n; ++i)

  {

   for(j = 0; j
<= m; ++j)

   {

    if(j
>= a[i].f1) dp[i+1][ j-a[i].f1 ] = min(dp[i+1][
j-a[i].f1 ] , dp[i][j]+a[i].t1); //fast

    dp[i+1][j]
= min(dp[i+1][j] , dp[i][j]+a[i].t2); //normal

    if(j
+ a[i].f2 <= m) //slow

     dp[i+1][
j+a[i].f2 ] = min(dp[i+1][ j+a[i].f2 ] , dp[i][j]+a[i].t3);

    else

     dp[i+1][m]
= min(dp[i+1][m] , dp[i][j]+a[i].t3);

   }

  }

  int result = INF;

  for(i = 0; i <=
m; ++i)

   result =
min(result,dp
[i]);

  printf("%d\n",result);

 }

 return 0;

}