标题: sicily_1024 Magic Island [打印本页] 作者: look_w 时间: 2019-2-18 19:28 标题: sicily_1024 Magic Island
题目
Constraints
Time Limit: 1 secs, Memory Limit: 32 MB
Description
There are N cities and N-1 roads in Magic-Island. You can go from one city to any other. One road only connects two cities. One day, The king of magic-island want to visit the island from the capital. No road is visited twice. Do you know the longest distance the king can go.
Input
There are several test cases in the input
A test case starts with two numbers N and K. (1<=N<=10000, 1<=K<=N). The cities is denoted from 1 to N. K is the capital.
The next N-1 lines each contain three numbers X, Y, D, meaning that there is a road between city-X and city-Y and the distance of the road is D. D is a positive integer which is not bigger than 1000.
Input will be ended by the end of file.
Output
One number per line for each test case, the longest distance the king can go.
Sample Input
struct Road {
int id_; // road id.
int end_; // the other end of the road
int length_; // the length of the road
Road(int id, int end, int length) {
id_ = id;
end_ = end;
length_ = length;
}
};
int maxDistance;
vector<Road> Roads[MAX_SIZE + 1]; // Roads[0] is useless
bool visited[MAX_SIZE + 1]; // visited[0] is also useless
// DFS searching the roads. for the implementation is recursive, the closed list
// have to be global.
// @Param start: a city that begin the visit.
// @Param pathLength: gets a certain path's length. if the path is longer than
// the max distance we knows, update the maxDistance.
void DFS(int start, int pathLength = 0) {
// for each road connected to start
for (int i = 0; i < Roads[start].size(); i++) {
// if the road is not visited
if (!visited[Roads[start].id_]) {
// visits it
visited[Roads[start].id_] = true;
// adds the road length to the path length
pathLength += Roads[start].length_;
// update the max distance
if (pathLength > maxDistance) maxDistance = pathLength;
// continue the path from the other end of the road
DFS(Roads[start].end_, pathLength);