zoukankan      html  css  js  c++  java
  • ZOJ----3471Most powerful(简单状压dp)

    题目

    Most Powerful
    Time Limit: 2 Seconds Memory Limit: 65536 KB
    Recently, researchers on Mars have discovered N powerful atoms. All of them are different. These atoms have some properties. When two of these atoms collide, one of them disappears and a lot of power is produced. Researchers know the way every two atoms perform when collided and the power every two atoms can produce.
    You are to write a program to make it most powerful, which means that the sum of power produced during all the collides is maximal.
    Input
    There are multiple cases. The first line of each case has an integer N (2 <= N <= 10), which means there are N atoms: A1, A2, … , AN. Then N lines follow. There are N integers in each line. The j-th integer on the i-th line is the power produced when Ai and Aj collide with Aj gone. All integers are positive and not larger than 10000.
    The last case is followed by a 0 in one line.
    There will be no more than 500 cases including no more than 50 large cases that N is 10.
    Output
    Output the maximal power these N atoms can produce in a line for each case.
    Sample Input
    2
    0 4
    1 0
    3
    0 20 1
    12 0 1
    1 10 0
    0
    Sample Output
    4
    22

    大意:有很多种气体(n<=10),任意两种气体可以相互反应,消耗掉其中一种气体并释放出一定的能量(消耗两种气体释放的能量不一样),求最大释放的能量有多组数据

    随便写写的水题,dp[state]表示当前状态为state时的最大消耗,枚举还没被消耗的气体就可以转移状态了

    #include<bits/stdc++.h>
    using namespace std;
    int dp[2000],n,p[12][12];
    int main(){
        	while(scanf("%d",&n),n)
        	{
            		memset(dp,0,sizeof(dp));
            		for(int i=1;i<=n;i++)
            		{
        	        		for(int j=1;j<=n;j++)
    	            		{
    	                			scanf("%d",&p[i][j]);
    	            		}
    	        	}
    	        	int state=1<<n;
    	        	for(int i=0;i<state;i++)
    	    	    {
    	        	    	for(int j=1;j<=n;j++)
    	        		    {
    		                		if(i&(1<<(j-1))) continue;
    	                			for(int k=1;k<=n;k++)
                	    			{
    		                    			if(j==k) continue;
    		                    			if(i&(1<<(k-1))) continue;
                        					dp[i+(1<<(k-1))]=max(dp[i+(1<<(k-1))],dp[i]+p[j][k]);
                    				}
                			}
    	        	}
        	    	int ans=0;
    	        	for(int i=0;i<state;i++)
    	        	{
    	            		ans=max(ans,dp[i]);
            		}
            		printf("%d
    ",ans);
            	}
        	return 0;
    }
    
  • 相关阅读:
    创建共享内存函数CreateFileMapping()详解
    窗口类、窗口类对象与窗口
    ubuntu中文版切换为英文后字体变化问题解决
    安装ubuntu12.04LTS卡住以及花屏问题
    时钟周期、振荡周期、机器周期、CPU周期、状态周期、指令周期、总线周期、任务周期
    波特率
    myod
    mycp
    20165226 2017-2018-2《Java程序设计》课程总结
    2017-2018-2 20165226 实验五《网络编程与安全》实验报告
  • 原文地址:https://www.cnblogs.com/forever-/p/9736072.html
Copyright © 2011-2022 走看看