ACM International Collegiate Programming H.动态规划

最新推荐文章于 2020-05-26 21:26:15 发布

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最新推荐文章于 2020-05-26 21:26:15 发布

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分类专栏：动态规划文章标签：动态规划

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本文介绍了一个三维空间中，从特定起点出发，通过有限的移动方式到达终点过程中收集最多“战利品”的问题。该问题采用动态规划解决，通过逆向思维从目标点反推至起点，实现最大值累积。

H - Commandos

commandos.in / standard output
Statements

A commando is a soldier of an elite light infantry often specializing in amphibious landings, abseiling or parachuting. This time our Commando unit wants to free as many hostages as it could from a hotel in Byteland, This hotel contains 10 identical floors numbered from 1 to 10 each one is a suite of 10 by 10 symmetric square rooms, our unit can move from a room (F, Y, X) to the room right next to it (F, Y, X + 1) or front next to it (F, Y + 1, X) and it can also use the cooling system to move to the room underneath it (F - 1, Y, X).

Knowing that our unit parachuted successfully in room 1-1 in floor 10 with a map of hostages locations try to calculate the maximum possible hostages they could save.

Input

Your program will be tested on one or more test cases. The first line of the input will be a single integer T. Followed by the test cases, each test case contains a number N (1 ≤ N ≤ 1, 000) representing the number of lines that follows. Each line contains 4 space separated integers (1 ≤ F, Y, X, H ≤ 10) means in the floor number F room Y-X there are H hostages.

Output

For each test case, print on a single line, a single number representing the maximum possible hostages that they could save.

Example

Input

Output

10
8

题意：一个三维空间，从（10，1，1）开始走，每次只能向下、向右或者向前走，求走到（1，10，10）的最大和的值。

解析：这个题目是个典型的动态规划题。类似于动态三角形，这里是个三维的。只需要处理边界情况就行了。

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;

int dp[11][11][11],a[11][11][11];
int main()
{
    freopen("commandos.in","r",stdin);
    int F,x,y,z,T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));
        scanf("%d",&n);
        for(int i=1; i<=n; ++i)
        {
            scanf("%d%d%d%d",&F,&x,&y,&z);
            a[F][x][y]=z;
        }
        for(int i=10; i>=1; i--)
            for(int j=1; j<=10; j++)
                for(int k=1; k<=10; k++)
                {
                    if(i!=10) dp[i][j][k]=max(dp[i][j][k],dp[i+1][j][k]);
                    if(j!=1) dp[i][j][k]=max(dp[i][j][k],dp[i][j-1][k]);
                    if(k!=1) dp[i][j][k]=max(dp[i][j][k],dp[i][j][k-1]);
                    dp[i][j][k]+=a[i][j][k];
                }
        printf("%d\n",dp[1][10][10]);
    }
    return 0;
}