HDOJ 题目5067 Harry And Dig Machine(状压dp,TSP)
来源:互联网 发布:访问js对象的属性 编辑:程序博客网 时间:2024/06/05 05:38
Harry And Dig Machine
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 806 Accepted Submission(s): 315
Problem Description
As we all know, Harry Porter learns magic at Hogwarts School. However, learning magical knowledge alone is insufficient to become a great magician. Sometimes, Harry also has to gain knowledge from other certain subjects, such as language, mathematics, English, and even algorithm.
Dumbledore, the headmaster of Hogwarts, is planning to construct a new teaching building in his school. The area he selects can be considered as an n*m grid, some (but no more than ten) cells of which might contain stones. We should remove the stones there in order to save place for the teaching building. However, the stones might be useful, so we just move them to the top-left cell. Taking it into account that Harry learned how to operate dig machine in Lanxiang School several years ago, Dumbledore decides to let him do this job and wants it done as quickly as possible. Harry needs one unit time to move his dig machine from one cell to the adjacent one. Yet skilled as he is, it takes no time for him to move stones into or out of the dig machine, which is big enough to carry infinite stones. Given Harry and his dig machine at the top-left cell in the beginning, if he wants to optimize his work, what is the minimal time Harry needs to finish it?
Dumbledore, the headmaster of Hogwarts, is planning to construct a new teaching building in his school. The area he selects can be considered as an n*m grid, some (but no more than ten) cells of which might contain stones. We should remove the stones there in order to save place for the teaching building. However, the stones might be useful, so we just move them to the top-left cell. Taking it into account that Harry learned how to operate dig machine in Lanxiang School several years ago, Dumbledore decides to let him do this job and wants it done as quickly as possible. Harry needs one unit time to move his dig machine from one cell to the adjacent one. Yet skilled as he is, it takes no time for him to move stones into or out of the dig machine, which is big enough to carry infinite stones. Given Harry and his dig machine at the top-left cell in the beginning, if he wants to optimize his work, what is the minimal time Harry needs to finish it?
Input
They are sever test cases, you should process to the end of file.
For each test case, there are two integers n and m.(1≤n,m≤50) .
The next n line, each line contains m integer. The j-th number ofith line a[i][j] means there are a[i][j] stones on the jth cell of the ith line.( 0≤a[i][j]≤100 , and no more than 10 of a[i][j] will be positive integer).
For each test case, there are two integers n and m.
The next n line, each line contains m integer. The j-th number of
Output
For each test case, just output one line that contains an integer indicate the minimal time that Harry can finish his job.
Sample Input
3 30 0 00 100 00 0 02 21 11 1
Sample Output
44
Source
BestCoder Round #14
Recommend
heyang | We have carefully selected several similar problems for you: 5604 5603 5602 5601 5600
题目大意:一个矩阵每走一个相邻的格子消耗一个时间,每个矩阵里边大于0的地方都是石头,他要从左上角走把所有石头都走一遍,然后再回到左上角
ac代码
#include<stdio.h>#include<string.h>#include<stdlib.h>#include<iostream>#include<algorithm>#define INF 0x3f3f3f3f#define mod 1000000007using namespace std;int dp[(1<<10)+10][15];int map[110][110],dis[110][110];int id[110][110];struct s{ int x,y;}a[150];int main(){ int n,m; while(scanf("%d%d",&n,&m)!=EOF) { int i,j,k; memset(id,0,sizeof(id)); // memset(dis,INF,sizeof(dis)); int cc=0; for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { scanf("%d",&map[i][j]); if(map[i][j]>0) { a[cc].x=i; a[cc].y=j; id[i][j]=cc; cc++; } } } for(i=0;i<cc;i++) { for(j=0;j<cc;j++) { dis[i][j]=abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y); } } memset(dp,INF,sizeof(dp)); for(i=0;i<cc;i++) { dp[1<<i][i]=abs(1-a[i].x)+abs(1-a[i].y); } int rt=(1<<cc)-1; for(i=1;i<=rt;i++) { for(j=0;j<cc;j++) { if(dp[i][j]==INF) continue; for(k=0;k<cc;k++) { if(dis[j][k]==INF||(i&(1<<k))) continue; int st=i|(1<<k); dp[st][k]=min(dp[st][k],dp[i][j]+dis[j][k]); } } } int ans=INF; for(i=0;i<cc;i++) { if(ans>dp[rt][i]+abs(1-a[i].x)+abs(1-a[i].y)) ans=dp[rt][i]+abs(1-a[i].x)+abs(1-a[i].y); } if(ans==INF) ans=0; printf("%d\n",ans); }}
0 0
- HDOJ 题目5067 Harry And Dig Machine(状压dp,TSP)
- HDU 5067 Harry And Dig Machine(状压DP)(TSP问题)
- HDOJ 5067 Harry And Dig Machine 状压DP
- hdu5067 Harry And Dig Machine(裸TSP问题 状压dp )
- HDU 5067 Harry And Dig Machine(状压dp)
- HDU-5067-Harry And Dig Machine(状压DP)
- HDOJ 5067Harry And Dig Machine(状态压缩DP)
- 【BestCoder】 HDOJ 5067 Harry And Dig Machine
- HDOJ 5067 Harry And Dig Machine
- hdu 5067 Harry And Dig Machine(dp)
- HDU 5067 Harry And Dig Machine(状压dp)
- hdu 5067 Harry And Dig Machine(状态压缩dp)
- hdu 5067 Harry And Dig Machine (TSP类,简单)
- hdu 5067 Harry And Dig Machine(状压DP,旅行商问题)
- HDU 5067 Harry And Dig Machine 状态压缩 DP
- hdu 5067 Harry And Dig Machine (状态压缩dp)
- BestCoder Round #14 B 题 Harry And Dig Machine 【TSP】
- Harry And Dig Machine
- Linux 动态库,静态库
- java 学习资料
- Android学习笔记之自定义View(钢琴键盘部件)
- vbs操作Excel生成lua配置文件
- 基于FBX SDK的FBX模型解析与加载 -(一)
- HDOJ 题目5067 Harry And Dig Machine(状压dp,TSP)
- 【数据结构与算法分析】证logX<X 对所有的X>0成立
- 关于:javaScript正则表达式语法
- LeetCode 091 Decode Ways
- 线性表链式存储结构下基本操作的实现(初始化、赋值、取值、插入、删除、归并等)
- android优秀开源代码
- Qt开发程序在Windows 10应用需要管理员运行的解决思路
- GLSL学习资料
- 开启从菜鸟走向大数据高手之路