Uva - 116 - Unidirectional TSP(多段图的最短路,动态规划)
来源:互联网 发布:ubuntu安装win7系统 编辑:程序博客网 时间:2024/05/17 07:26
Background
Problems that require minimum paths through some domain appear in many different areas of computer science. For example, one of the constraints in VLSI routing problems is minimizing wire length. The Traveling Salesperson Problem (TSP) -- finding whether all the cities in a salesperson's route can be visited exactly once with a specified limit on travel time -- is one of the canonical examples of an NP-complete problem; solutions appear to require an inordinate amount of time to generate, but are simple to check.
This problem deals with finding a minimal path through a grid of points while traveling only from left to right.
The Problem
Given an matrix of integers, you are to write a program that computes a path of minimal weight. A path starts anywhere in column 1 (the first column) and consists of a sequence of steps terminating in column n (the last column). A step consists of traveling from column i to column i+1 in an adjacent (horizontal or diagonal) row. The first and last rows (rows 1 and m) of a matrix are considered adjacent, i.e., the matrix ``wraps'' so that it represents a horizontal cylinder. Legal steps are illustrated below.
The weight of a path is the sum of the integers in each of the n cells of the matrix that are visited.
For example, two slightly different matrices are shown below (the only difference is the numbers in the bottom row).
The minimal path is illustrated for each matrix. Note that the path for the matrix on the right takes advantage of the adjacency property of the first and last rows.
The Input
The input consists of a sequence of matrix specifications. Each matrix specification consists of the row and column dimensions in that order on a line followed by integers where m is the row dimension and n is the column dimension. The integers appear in the input in row major order, i.e., the first n integers constitute the first row of the matrix, the second n integers constitute the second row and so on. The integers on a line will be separated from other integers by one or more spaces. Note: integers are not restricted to being positive. There will be one or more matrix specifications in an input file. Input is terminated by end-of-file.
For each specification the number of rows will be between 1 and 10 inclusive; the number of columns will be between 1 and 100 inclusive. No path's weight will exceed integer values representable using 30 bits.
The Output
Two lines should be output for each matrix specification in the input file, the first line represents a minimal-weight path, and the second line is the cost of a minimal path. The path consists of a sequence of n integers (separated by one or more spaces) representing the rows that constitute the minimal path. If there is more than one path of minimal weight the path that islexicographically smallest should be output.
Sample Input
5 63 4 1 2 8 66 1 8 2 7 45 9 3 9 9 58 4 1 3 2 63 7 2 8 6 45 63 4 1 2 8 66 1 8 2 7 45 9 3 9 9 58 4 1 3 2 63 7 2 1 2 32 29 10 9 10
Sample Output
1 2 3 4 4 5161 2 1 5 4 5111 119
思路:题目要求从左边到右边的最小权路径。输每列出行号。逆推求解。
1.设dp[i][j]为 从(i,j)位置出发,到最右边的最小权路径的权值。
2.状态转移方程:当前位置的最小权路径权值为,当前位置的权值+ 下一列的同一行,上一行和下一行的最小值。
3.边界为最后一行,从最后一行逆推到第一行。(注意第一行和最后一行的处理)
4.找出第一行的dp[][]最小值。
5.记得记录路径。
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int d[11][101]; int nextpath[11][101]; //记录路径 int matrix[11][101];int n,m,first;int main(){while(scanf("%d%d",&n,&m)!=EOF){//读取数据 for(int i=0 ;i<n ;i++){for(int j=0 ;j<m ;j++){scanf("%d",&matrix[i][j]);}} //逆推过程 for(int j=m-1 ;j>=0 ;j--){for(int i=0 ;i<n ;i++){if(j==(m-1))d[i][j] = matrix[i][j];else{int row[3] = {i, i-1 ,i+1}; if(i==0) row[1] = n-1;if(i==(n-1)) row[2] = 0;sort(row,row+3);d[i][j] = 100000;int temp = matrix[i][j]; //状态转换方程 for(int k=0 ;k<3 ;k++){ if(d[i][j]>(temp+d[row[k]][j+1])){d[i][j] = (temp+d[row[k]][j+1]);nextpath[i][j] = row[k];}} }} }//寻找最大值 int ans = 1000000000;for(int i=0 ;i<n ;i++){if(ans>d[i][0]){ans = d[i][0];first = i+1;}}//输出路径 printf("%d",first);for(int i=nextpath[first-1][0],j=1 ;j<m ; i=nextpath[i][j],j++){printf(" %d",i+1);}printf("\n%d\n",ans);}return 0;}
- Uva - 116 - Unidirectional TSP(多段图的最短路,动态规划)
- 动态规划--多段图的最短路--单向TSp(Unidirectional TSP ,UVa 116)
- UVA - 116 Unidirectional TSP(dp+多段图的最短路)
- UVA - 116 Unidirectional TSP 多段图的最短路
- UVa 116 Unidirectional TSP (多段图的最短路 DP)
- UVA - 116 - Unidirectional TSP(动态规划)
- uva 116 - Unidirectional TSP (动态规划)
- 动态规划(Unidirectional TSP,uva 116)
- UVA 116 Unidirectional TSP(动态规划)
- 【uva】116-Unidirectional TSP(动态规划,路径问题)
- UVA 116 Unidirectional TSP 动态规划+类似数塔
- [中等] UVa OJ 116 Unidirectional TSP 动态规划
- uva116-Unidirectional TSP (动态规划,多段图)
- UVa 116 Unidirectional TSP
- UVA 116 - Unidirectional TSP
- UVa 116 - Unidirectional TSP
- uva 116 Unidirectional TSP
- UVa 116Unidirectional TSP
- telnet两三事
- 数组
- Android-图片处理Picasso库的解析使用
- [web] Haproxy配置
- linux 的线程
- Uva - 116 - Unidirectional TSP(多段图的最短路,动态规划)
- static
- mysqldump,mydumper以及xtrabackup备份流程简述
- 源码推荐:快速集成仿QQ弹出菜单 流畅的转场实现 选座实现
- 三人借书(5本)问题
- windows学习历程-IPC之互斥对象
- Spark快速大数据分析-RDD编程
- Xenomai 2.6.5 on Ubuntu 14.04
- 数制转换(C语言版)