programming-challenges Unidirectional TSP (111104) 题解

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http://algorithmist.com/index.php/UVa_116 这里有很好的测试用例。这道题其实只是很直观的迭代。

#include <iostream>#include <sstream>#include <fstream>#include <string>#include <vector>#include <list>#include <queue>#include <map>#include <set>#include <stack>#include <assert.h>#include <algorithm>#include <math.h>#include <ctime>#include <functional>#include <string.h>#include <stdio.h>#include <numeric>#include <float.h>using namespace std;const int N = 110, M = 20; long long a[M][N]; struct Rec {vector<int> path; };bool operator<(const Rec& rec1, const Rec& rec2) {for (int i = 0; i < rec1.path.size(); i++) {if (rec1.path[i] < rec2.path[i]) return true;if (rec1.path[i] > rec2.path[i]) return false;}return false; }vector<Rec> paths(M); int main() {int m, n; while (cin >> m) {cin >> n; memset(a, 0, sizeof(a));paths.clear();paths.resize(M);for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {cin >> a[i][j]; }}for (int i = 0; i < m; i++) {paths[i].path.push_back(i);}for (int i = 1; i < n; i++) {vector<Rec> tempPaths(M);for (int j = 0; j < m; j++) {long long minVal = a[(j + m - 1) % m][i-1] + a[j][i];int row = (j + m - 1) % m; for (int k = 0; k <= 1; k++) {long long val = a[(j + m + k) % m][i-1] + a[j][i];if (val < minVal) {minVal = val; row = (j + m + k) % m; }else if (val == minVal) {int newRow = (j + m + k) % m; if (paths[row].path > paths[newRow].path) {row = newRow; } }}tempPaths[j] = paths[row];tempPaths[j].path.push_back(j); a[j][i] = minVal; }paths = tempPaths;}long long obj = 0x7fffffffffffffff; for (int i = 0; i < m; i++) {obj = obj < a[i][n - 1] ? obj : a[i][n - 1];}bool firstPath = true;Rec bestPath; for (int i = 0; i < m; i++) {if (a[i][n - 1] == obj) {if (firstPath) {firstPath = false;bestPath = paths[i];}else {bestPath = min(bestPath, paths[i]);}}}for (int i = 0; i < bestPath.path.size(); i++) {cout << bestPath.path[i] + 1;if (i + 1 < bestPath.path.size())cout << " ";}cout << endl << obj << endl; }return 0; }

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