Problem Links:
uva00116,Problem:
Unidirectional TSP
Unidirectional TSP |
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

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


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
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 is lexicographically smallest should be output.Sample Input
5 6 3 4 1 2 8 6 6 1 8 2 7 4 5 9 3 9 9 5 8 4 1 3 2 6 3 7 2 8 6 4 5 6 3 4 1 2 8 6 6 1 8 2 7 4 5 9 3 9 9 5 8 4 1 3 2 6 3 7 2 1 2 3 2 2 9 10 9 10
Sample Output
1 2 3 4 4 5 16 1 2 1 5 4 5 11 1 1 19
Solution:
Using dynamic programming: consume[i][j] = min{consume[i][j+1], consume[i-1][j+1], consume[i+1][j+1]} + matrix[i][j];
PS: Computer from right to left will result lexicographically.
Backtracking might be the optional way.
Test cases:
Input 5 6 3 4 1 2 8 6 6 1 8 2 7 4 5 9 3 9 9 5 8 4 1 3 2 6 3 7 2 8 6 4 5 6 3 4 1 2 8 6 6 1 8 2 7 4 5 9 3 9 9 5 8 4 1 3 2 6 3 7 2 1 2 3 2 2 9 10 9 10 12 14 1 2 2 1 1 1 1 1 1 1 1 1 1 2 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 4 7 1 2 -3 4 -2 1 5 -1 3 5 -2 6 -3 4 2 1 3 -2 -1 3 1 3 -3 4 2 -3 4 3 5 6 3 4 1 2 8 6 6 1 8 2 7 4 5 9 3 9 9 5 8 4 1 3 2 6 3 7 2 8 6 4 5 6 3 4 1 2 8 6 6 1 8 2 7 4 5 9 3 9 9 5 8 4 1 3 2 6 3 7 2 1 2 3 2 2 9 10 9 10 5 6 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5 3 4 1 2 3 4 1 2 3 4 1 2 3 4 5 5 1 5 10 6 3 5 1 8 4 11 10 12 5 2 9 7 3 20 5 8 4 1 5 12 6 5 10 11 53 34 73 18 53 99 52 31 54 4 72 24 6 46 17 63 82 89 25 67 22 10 97 99 64 33 45 81 76 24 71 46 62 18 11 54 40 17 51 99 8 57 76 7 51 90 92 51 21 5 10 11 53 1 73 18 53 99 52 31 54 4 72 54 6 46 17 63 82 89 25 67 22 80 97 99 64 33 45 81 76 24 71 46 62 18 11 54 40 17 51 99 8 57 76 7 51 90 92 51 21 5 6 -3 -4 -1 -2 -8 -6 -6 -1 -8 -2 -7 -4 -5 -9 -3 -9 -9 -5 -8 -4 -1 -3 -2 -6 -3 -7 -2 -8 -6 -4 10 100 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 29 1 -1 0 0 1 -1 -1 -1 1 1 0 1 -1 -1 -1 0 -1 -1 1 -1 1 0 0 -1 0 -1 1 1 0 -1 1 0 -1 -1 0 1 0 -1 -1 -1 -1 1 1 0 -1 -1 0 -1 -1 1 -1 1 0 -1 1 0 1 -1 0 0 1 0 -1 1 0 0 1 1 0 -1 1 1 -1 -1 1 0 0 1 1 0 -1 1 0 -1 -1 1 1 1 -1 -1 -1 -1 -1 0 0 0 1 -1 0 0 0 -1 -1 0 -1 0 1 0 -1 1 0 1 1 -1 0 1 Output 1 2 3 4 4 5 16 1 2 1 5 4 5 11 1 1 19 1 2 3 4 5 6 7 8 9 10 11 12 1 2 14 1 4 1 2 1 2 3 -11 1 2 3 4 4 5 16 1 2 1 5 4 5 11 1 1 19 1 1 1 1 1 1 6 1 1 1 1 10 1 2 3 2 1 14 2 3 3 2 1 2 3 4 4 5 188 1 5 1 2 1 2 3 4 4 5 172 4 3 2 3 3 4 -49 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 100 2 1 4 4 3 4 1 1 2 2 2 2 1 1 1 2 1 1 2 1 4 4 1 1 2 1 4 1 2 -25
Source Code:
//Sun Apr 10 03:06:20 CDT 2011#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define INT_MAX 2147483647
#define INT_MIN -2147483647
using namespace std;
class Point {
public:
int x;
int y;
};
void init(int m, int n, vector<vector<int> > &matrix,
vector<vector<Point> > &parent) {
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++) {
cin >> matrix[i][j];
parent[i][j].x = i;
parent[i][j].y = j;
}
// for (int i = 0; i < m; i++) {
// for (int j = 0; j < n; j++)
// cout << matrix[i][j] << " ";
// cout << endl;
// }
}
void reverseVector(vector<vector<int> > &v) {
for (int i = 0; i < (int) v.size(); i++)
reverse(v[i].begin(), v[i].end());
}
bool cmp(const vector<int> A, const vector<int> B) {
for (int i = 0; i < (int) A.size(); i++) {
if (A[i] > B[i])
return false;
if (A[i] < B[i])
return true;
}
return true;
}
pair<int, vector<int> > Path(int m, int n, vector<vector<int> > &matrix,
vector<vector<Point> > &parent) {
vector<vector<int> > consume(m, vector<int> (n, INT_MAX));
// for (int i = 0; i < m; i++) {
// for (int j = 0; j < n; j++)
// cout << consume[i][j] << " ";
// cout << endl;
// }
for (int i = 0; i < m; i++)
consume[i][n - 1] = matrix[i][n - 1];
//consume[0][0] = matrix[0][0];
for (int j = n - 2; j >= 0; j--) {
for (int i = 0; i < m; i++) {
vector<int> pos;
pos.push_back(i);
pos.push_back((i + 1) % m);
pos.push_back((i + m - 1) % m);
sort(pos.begin(), pos.end());
for (int k = 0; k < 3; k++) {
if (consume[pos[k]][j + 1] != INT_MAX && consume[i][j]
> matrix[i][j] + consume[pos[k]][j + 1]) {
consume[i][j] = matrix[i][j] + consume[pos[k]][j + 1];
parent[i][j].x = pos[k];
parent[i][j].y = j + 1;
}
}
}
}
// for (int j = 0; j < n; j++)
// for (int i = 0; i < m; i++) {
// cout << consume[i][j] << ": " << parent[i][j].x << ", "
// << parent[i][j].y << endl;
// }
// for (int i = 0; i < m; i++) {
// for (int j = 0; j < n; j++)
// cout << consume[i][j] << " ";
// cout << endl;
// }
int mmin = INT_MAX;
for (int i = 0; i < m; i++) {
if (mmin > consume[i][0]) {
mmin = consume[i][0];
}
}
vector<vector<int> > ret;
for (int i = 0; i < m; i++) {
if (consume[i][0] == mmin)
ret.push_back(vector<int> (1, i));
}
for (int j = 0; j < (int) ret.size(); j++) {
int xx = ret[j][0];
int yy = 0;
ret[j][0]++;
for (int i = 0; i < n - 1; i++) {
xx = parent[xx][yy].x;
yy = parent[xx][yy].y;
ret[j].push_back(xx + 1);
}
}
// reverseVector(ret);
// for (int i = 0; i < ret.size(); i++) {
// for (int j = 0; j < ret[i].size(); j++)
// cout << ret[i][j] << " ";
// cout << endl;
// }
sort(ret.begin(), ret.end(), cmp);
// for (int i = 0; i < ret.size(); i++) {
// for (int j = 0; j < ret[i].size(); j++)
// cout << ret[i][j] << " ";
// cout << endl;
// }
return make_pair(mmin, ret[0]);
}
void print(pair<int, vector<int> > v) {
cout << v.second[0];
for (int i = 1; i < (int) v.second.size(); i++)
cout << " " << v.second[i];
cout << endl;
cout << v.first << endl;
}
int main(int argc, char* argv[]) {
//freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
int m, n;
while (cin >> m >> n) {
vector<vector<int> > matrix(m, vector<int> (n, 0));
vector<vector<Point> > parent(m, vector<Point> (n));
init(m, n, matrix, parent);
print(Path(m, n, matrix, parent));
}
//fclose(stdin);
//fclose(stdout);
return 0;
}
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