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\nMinimum Number of Trips = ", count);
if (remainder)
printf("%d\n", number_customer / max_value + 1);
else
printf("%d\n", number_customer / max_value);
// 6. 读取下一行数据
scanf("%d%d", &number_points, &number_ways);
count++;
}
return 0;
} |