CFGym - 101147H. Commandos 动态规划

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在一个10x10x10的建筑中，特种部队需要在限定时间内从顶层开始，通过移动到相邻房间来解救人质。任务是计算能解救的最大人质数量。

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1.题目描述：

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

2.题意概述：

在一个10x10x10的楼里面有人质在不同楼层不同方位，特种队员在 (F, Y, X)单位时

间可以向(F, Y, X + 1)或者(F, Y + 1, X)或者(F - 1, Y, X)方向走，同时解救该方向人质，

问你最多可以解救多少人质

3.解题思路：

既可以搜索，也可以dp，我当时做法是dp，转移方程

是dp[i][j][k] = max(dp[i][j][k], max(dp[i - 1][j][k], max(dp[i][j - 1][k], dp[i][j][k - 1]))) + a[i][j][k]; }

注意的是自顶部向下走，则要对楼层进行颠倒

4.Ac代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <ctime>
#define N 11
using namespace std;
int a[N][N][N];
int dp[N][N][N];

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
        long _begin_time = clock();
    #endif
    freopen("commandos.in", "r", stdin);
	int t, n;
	scanf("%d", &t);
	while (t--)
	{
        memset(a, 0, sizeof(a));
		memset(dp, 0, sizeof(dp));
		scanf("%d", &n);
		for (int i = 0; i < n; i++)
		{
            int f, y, x, h;
            scanf("%d%d%d%d", &f, &y, &x, &h);
            a[11 - f][y][x] = h;
		}
        for (int i = 1; i <= 10; i++)
        {
            for (int j = 1; j <= 10; j++)
            {
                for (int k = 1; k <= 10; k++)
                {
                    dp[i][j][k] = max(dp[i][j][k], max(dp[i - 1][j][k], max(dp[i][j - 1][k], dp[i][j][k - 1]))) + a[i][j][k];
                }
            }
        }
        printf("%d\n", dp[10][10][10]);
	}
    #ifndef ONLINE_JUDGE
        long _end_time = clock();
        printf("time = %ld ms\n", _end_time - _begin_time);
    #endif
    return 0;
}