10099 - The Tourist Guide
题目页：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1040
打不开的可以上国内的：http://acm.sdibt.edu.cn/JudgeOnline/problem.php?id=1080
之前使用 最小生成树的 Kruskal 算法 + 广度优先搜索
改进之后用了 用动态规划技术实现的所有节点对的最短路径问题的 Floyd-Warshall 算法，不仅代码量急剧减少，简化了逻辑
此题的坑之一：Mr.G.自己也算一个人……… 反映到代码里是63行
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#include <stdio.h>#include <stdlib.h>int number_points;int map_values[110][110];void floyd(){ for (int a = 1; a <= number_points; a++) { for (int b = 1; b <= number_points; b++) { for (int x = 1; x <= number_points; x++) { int ax = map_values[a][x]; int xb = map_values[x][b]; int smaller = ax < xb ? ax : xb; if (map_values[a][b] < smaller) { map_values[a][b] = smaller; } } } }}int main(){ int count = 1; int number_ways; scanf("%d%d", &number_points, &number_ways); while(number_points != 0 && number_ways != 0) { // 0. 初始化图 for (int y = 1; y <= number_points; y++) for (int x = 1; x <= number_points; x++) { map_values[x][y] = 0; } // 1. 读取边 for (int i = 0; i < number_ways; i++) { int x, y, values; scanf("%d%d%d", &x, &y, &values); map_values[x][y] = values; map_values[y][x] = values; } // 2. 读取起始节点和顾客数量 int start, end, number_customer; scanf("%d%d%d", &start, &end, &number_customer); // 3. floyd 算法 floyd(); // 4. 获取值 int max_value = map_values[start][end]; // 5. 输出结果 max_value--; // Mr. G. 自己也算上 int remainder = number_customer % max_value; printf("Scenario #%d
Minimum Number of Trips = ", count); if (remainder) printf("%d
", number_customer / max_value + 1); else printf("%d
", number_customer / max_value); // 6. 读取下一行数据 scanf("%d%d", &number_points, &number_ways); count++; } return 0;} |