hdu 3920(状态压缩dp)

最新推荐文章于 2019-12-18 20:07:00 发布

原创最新推荐文章于 2019-12-18 20:07:00 发布 · 395 阅读

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本文介绍了一个关于使用激光枪射击敌人并优化射击路径的算法问题。玩家需清除地图上的所有敌人，通过激光反射机制同时攻击两个目标，旨在找到最节省能量的射击方案。文章探讨了状态压缩动态规划的应用，并给出了两种不同的实现方式。

Clear All of Them I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 122768/62768 K (Java/Others)

Problem Description

Acmers have been the Earth Protector against the evil enemy for a long time, now it’s your turn to protect our home.
  There are 2 * n enemies in the map. Your task is to clear all of them with your super laser gun at the fixed position (x, y).
  For each laser shot, your laser beam can reflect 1 times (must be 1 times), which means it can kill 2 enemies at one time. And the energy this shot costs is the total length of the laser path.
  For example, if you are at (0, 0), and use one laser shot kills the 2 enemies in the order of (3, 4), (6, 0), then the energy this shot costs is 5.0 + 5.0 = 10. 00.
  Since there are 2 * n enemies, you have to shot n times to clear all of them. For each shot, it is you that select two existed enemies and decide the reflect order.
  Now, telling you your position and the 2n enemies’ position, to save the energy, can you tell me how much energy you need at least to clear all of them?
  Note that:
   > Each enemy can only be attacked once.
   > All the positions will be unique.
   > You must attack 2 different enemies in one shot.
   > You can’t change your position.

Input

The first line contains a single positive integer T( T <= 100 ), indicates the number of test cases.
For each case:
  There are 2 integers x and y in the first line, which means your position.
  The second line is an integer n(1 <= n <= 10), denote there are 2n enemies.
  Then there following 2n lines, each line have 2 integers denote the position of an enemy.

  All the position integers are between -1000 and 1000.

Output

For each test case: output the case number as shown and then print a decimal v, which is the energy you need at least to clear all of them (round to 2 decimal places).

Sample Input

Sample Output

Case #1: 6.00
Case #2: 4.41


解题思路：首先这道题目最多只有20个状态，打中目标还有考虑别的目标是否打中，属于取与不取的问题，那么正好就是状态压缩能够解决的范畴。。dp[i]表示状态为i时的最小化费，那么一次能够打两个，那么下一个状态应该是i-(1<<j)-(1<<k)，j，k分别代表先后击中的目标。最后的目标是dp[0]（0代表该位置的目标被击中，1代表未被击中）。可惜最后TLE,后面仔细想想，应该还是在先被击中与后被击中的选择上循环次数太多啦。。。

TLE:
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;

const int maxn = 55;
const double esp = 1e-8;
struct node
{
	double x,y;
}o,s[maxn];
int n;
double dp[(1<<20)+5],d[maxn][maxn],os[maxn];

double dis(node a,node b)  
{  
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));  
}  

int main()
{	
	int t,i,j,k,cas = 1;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf%lf%d",&o.x,&o.y,&n);
		n *= 2;
		for(i = 0; i<n; i++)
			scanf("%lf%lf",&s[i].x,&s[i].y);
		for(i = 0; i<n; i++)
			for(j = 0; j<n; j++)
			{
				d[i][j] = dis(s[i],s[j]);
			}
			for(i = 0; i<n; i++)
				os[i] = dis(o,s[i]);
			for(i = 0; i<(1<<n); i++)
				dp[i] = -1.0;
			dp[(1<<n)-1] = 0;
			for(i = (1<<n)-1; i >= 0; i--)
				for(j = 0; j < n; j++)	//第一次打到j
					for(k = 0; k < n; k++)	//第二次达到k
					{
						if(dp[i] == -1) continue;
						if(i & (1<<j) && i & (1<<k) && j != k)
						{
							int tmp = i - (1<<j) - (1<<k);
							if(dp[tmp] < esp || dp[tmp] > dp[i] + os[j] + d[j][k])
								dp[tmp] = dp[i] + os[j] + d[j][k];
						}
					}
			printf("Case #%d: %.2f\n",cas++,dp[0]);
	}
	return 0;
}


看了别人的分析：所以状态的转移是由前往后递推的，假如当前状态是cur，上一个状态是last，应该满足存在i！=j有last&(1<<i)=0且last&(1<<j)=0且last | (1<<i)|(1<<j) = cur，由此来更新当前的状态cur，最后的答案就是DP[(1<<(2n)) - 1]
注意到上面我们是要枚举i和j的，所以这个的总复杂度就是20 * 20 * 2^20，这显然是会超时的， 所以需要优化
注意到如果存在两队人(a,b)(c,d)我们先打(a, b)再打(c, d)和先打（c,d）在打(a, b)是一样的，所以我们完全可以每次取last中最小的一位是0的与后面所有的0组合，这样不仅没有漏掉解，而且复杂度也将到了O(20 * 2^20)这就可以过了

AC:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define exp 1e-8

struct node
{
    double x,y;
} o,s[50];

int n;
double d[50][50],dp[(1<<20)+1],os[50];

double dis(node a,node b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

int cmp(node a,node b)
{
    return dis(a,o)<dis(b,o);
}

double dfs(int u)
{
    if(dp[u]>exp) return dp[u];
    if(!u) return 0;
    int m = 0;
    while(!(u&(1<<m))) m++;
    int tem;
    for(int i = m+1; i<n; i++)
    {
        if(u&(1<<i))
        {
            tem = u - (1<<i) - (1<<m);
            double ans = dfs(tem)+os[m]+d[i][m];
            if(dp[u]<exp || ans<dp[u])
                dp[u] = ans;
        }
    }
    return dp[u]<exp?0:dp[u];
}

int main()
{
    int t,i,j,k,cas = 1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%d",&o.x,&o.y,&n);
        n*=2;
        for(i = 0; i<n; i++)
            scanf("%lf%lf",&s[i].x,&s[i].y);
        sort(s,s+n,cmp);
        for(i = 0; i<n; i++)
            for(j = 0; j<n; j++)
            {
                d[i][j] = dis(s[i],s[j]);
            }
        for(i = 0; i<n; i++)
            os[i] = dis(o,s[i]);
        for(i = 0; i<(1<<n); i++)
            dp[i] = -1.0;
        dfs(i-1);
        printf("Case #%d: %.2f\n",cas++,dp[i-1]);
    }

    return 0;
}