ZOJ 3541

题目大意:有n个按钮,第i个按钮在按下ti 时间后回自动弹起,每个开关的位置是di,问什么策略按开关可以使所有的开关同时处于按下状态


Description


There is one last gate between the hero and the dragon. But opening the gate isn't an easy task.

There were n buttons list in a straight line in front of the gate and each with an integer on it. Like other puzzles the hero had solved before, if all buttons had been pressed down in any moment, the gate would open. So, in order to solve the puzzle, the hero must press all the button one by one.

After some trials, the hero found that those buttons he had pressed down would pop up after a while before he could press all the buttons down. He soon realized that the integer on the button is the time when the button would automatic pop up after pressing it, in units of second. And he measured the distance between every button and the first button, in units of maximum distance the hero could reach per second. Even with this information, the hero could not figure out in what order he should press the buttons. So you talent programmers, are assigned to help him solve the puzzle.

To make the puzzle easier, assuming that the hero always took integral seconds to go from one button to another button and he took no time turnning around or pressing a button down. And the hero could begin from any button.

Input


The input file would contain multiple cases. Each case contains three lines. Process to the end of file.

The first line contains a single integer n(1 ≤ n ≤200), the number of buttons.

The second line contains n integers T1, T2, ..., Tn, where Ti(1 ≤ Ti ≤ 1,000,000) is the time the ith button would automatic pop up after pressing it, in units of second.

The third line contains n integers D1, D2, ..., Dn, where Di(1 ≤ Di ≤ 1,000,000) is the time hero needed to go between the ith button and the first button, in units of second. The sequence will be in ascending order and the first element is always 0.

Output


Output a single line containing n integers which is the sequence of button to press by the hero. If there are multiply sequences, anyone will do. If there is no way for the hero to solve the puzzle, just output "Mission Impossible"(without quote) in a single line.

Sample Input


2
4 3
0 3
2
3 3
0 3
4
5 200 1 2
0 1 2 3

Sample Output


1 2
Mission Impossible
1 2 4 3

Hint


In the second sample, no matter which button the hero pressed first, the button would always pop up before he press the other button. So there is no way to make all the button pressed down.

Solution


本题很容易想到区间DP,对于一个区间,一定是从某个端点开始,因为如果从中间开始之后按别的开关时一定会经过这个点。

状态

\(f_{i,j,0/1}\)

表示区间[i,j]从左/右端点开始的最小时间。

状态转移方程见代码。

//Writer : Hsz %WJMZBMR%tourist%hzwer
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <cstdlib>
#include <algorithm>
const int inf=0x3fffffff;
#define LL long long
using namespace std;
const int N=222;
int n;
LL t[N],d[N],f[N][N][2];
bool way[N][N][2];
int main() {
while(scanf("%d",&n)!=EOF) {
memset(f,0,sizeof f);
for(int i=1; i<=n; i++)
scanf("%lld",&t[i]);
for(int i=1; i<=n; i++)
scanf("%lld",&d[i]);
for(int l=2; l<=n; l++) {
for(int i=1; i+l-1<=n; i++) {
int j=i+l-1; if(f[i+1][j][0]+d[i+1]-d[i]<f[i+1][j][1]+d[j]-d[i])
f[i][j][0]=f[i+1][j][0]+d[i+1]-d[i],way[i][j][0]=0; else f[i][j][0]=f[i+1][j][1]+d[j]-d[i],way[i][j][0]=1; if(t[i]<=f[i][j][0]||f[i][j][0]>=inf)
f[i][j][0]=inf; if(f[i][j-1][1]+d[j]-d[j-1]<=f[i][j-1][0]+d[j]-d[i])
f[i][j][1]=f[i][j-1][1]+d[j]-d[j-1],way[i][j][1]=1; else f[i][j][1]=f[i][j-1][0]+d[j]-d[i],way[i][j][1]=0; if(t[j]<=f[i][j][1]||f[i][j][1]>=inf)
f[i][j][1]=inf;
}
}
int l,r,v;
if(f[1][n][0]<inf) {
printf("1");
l=2,r=n,v=way[1][n][0];
} else if(f[1][n][1]<inf) {
printf("%d",n);
l=1,r=n-1,v=way[1][n][1];
} else {
puts("Mission Impossible");
continue;
}
while(l<=r) {
if(!v) printf(" %d",l),v=way[l][r][0],l++;
else printf(" %d",r),v=way[l][r][1],r--;
}
printf("\n");
}
return 0;
}

[ZOJ]3541 Last Puzzle (区间DP)的更多相关文章

  1. ZOJ 3469 Food Delivery 区间DP

    这道题我不会,看了网上的题解才会的,涨了姿势,现阶段还是感觉区间DP比较难,主要是太弱...QAQ 思路中其实有贪心的意思,n个住户加一个商店,分布在一维直线上,应该是从商店开始,先向两边距离近的送, ...

  2. POJ1651:Multiplication Puzzle(区间DP)

    Description The multiplication puzzle is played with a row of cards, each containing a single positi ...

  3. poj 1651 Multiplication Puzzle (区间dp)

    题目链接:http://poj.org/problem?id=1651 Description The multiplication puzzle is played with a row of ca ...

  4. zoj 3469 Food Delivery 区间dp + 提前计算费用

    Time Limit: 2 Seconds      Memory Limit: 65536 KB When we are focusing on solving problems, we usual ...

  5. ZOJ - 3469 Food Delivery (区间dp)

    When we are focusing on solving problems, we usually prefer to stay in front of computers rather tha ...

  6. ZOJ 3469 Food Delivery(区间DP好题)

    题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4255 题目大意:在x轴上有n个客人,每个客人每分钟增加的愤怒值不同. ...

  7. POJ 1651 Multiplication Puzzle 区间dp(水

    题目链接:id=1651">点击打开链 题意: 给定一个数组,每次能够选择内部的一个数 i 消除,获得的价值就是 a[i-1] * a[i] * a[i+1] 问最小价值 思路: dp ...

  8. POJ1651Multiplication Puzzle[区间DP]

    Multiplication Puzzle Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8737   Accepted:  ...

  9. ZOJ - 3537 Cake (凸包+区间DP+最优三角剖分)

    Description You want to hold a party. Here's a polygon-shaped cake on the table. You'd like to cut t ...

随机推荐

  1. HDU 3537

    翻硬币游戏,纯.. 注意要判重 #include <iostream> #include <cstdio> #include <cstring> #include ...

  2. cocos2d-android学习四 ---- 精灵的创建

    上篇文章我们创建了一个黑乎乎的界面.以下我们就给它增加一个精灵. 我们这次就一起来学习精灵的基础知识. 1.什么是精灵 游戏中全部会动的对象都是精灵,能够是主人公,背景元素,一个子弹或者是敌人. 一个 ...

  3. 【cl】selenium实例一:打开百度,获取第四个标题

    /*创建类的时候是TestNG Class*/ package Selenium_lassen; import static org.junit.Assert.*; import java.util. ...

  4. Online Object Tracking: A Benchmark 论文笔记

    Factors that affect the performance of a tracing algorithm 1 Illumination variation 2 Occlusion 3 Ba ...

  5. luogu3799 妖梦拼木棒

    题目大意 有n根木棒,现在从中选4根,想要组成一个正三角形,问有几种选法?木棒长度都<=5000. 题解 根据容斥原理,三角形两条边分别由长度相等的单根木棒组成,另一条边由两条小于该边长的木棒构 ...

  6. luogu1969 积木大赛

    题目大意 搭建一座宽度为n的大厦,大厦可以看成由n块宽度为1的积木组成,第i块积木的最终高度需要是hi. 在搭建开始之前,没有任何积木(可以看成n块高度为 0 的积木).接下来每次操作,可以选择一段连 ...

  7. [AHOI 2008] 聚会

    [题目链接] https://www.lydsy.com/JudgeOnline/problem.php?id=1832 [算法] 最近公共祖先 [代码] #include<bits/stdc+ ...

  8. AIX&nbsp;常用命令汇总(一)

    命令 内核 如何知道自己在运行 32 位内核还是 64 位内核? 要显示内核启用的是 32 位还是 64 位,可输入以下命令: bootinfo -K 如何知道自己在运行单处理器还是多处理器内核? / ...

  9. Solr.NET快速入门(三)【高亮显示】

    此功能会"高亮显示"匹配查询的字词(通常使用标记),包括匹配字词周围的文字片段. 要启用高亮显示,请包括HighlightingParameters QueryOptions对象, ...

  10. week2 notebook2

    Beginning day2: 1.基础数据类型宏观: 1.1.整型:int:1,2,3 1.2.字符串:str:‘anthony’ 1.2.1: 索引:索引即下标,就是字符串组成的元素从第一个开始, ...