Time Limit: 2000MS | Memory Limit: 30000K | |
Total Submissions: 4182 | Accepted: 2116 | |
Case Time Limit: 1000MS |
Description
Input
Output
Sample Input
7 6 1 6 13 E 6 3 9 E 3 5 7 S 4 1 3 N 2 4 20 W 4 7 2 S
Sample Output
52
Hint
第一道树的直径题
#include
#include
#include
#include
#include
#define maxn 40000+10
#define maxm 80000+10
#define INF 0x3f3f3f3f
using namespace std;
struct node{
int u, v, w, next;
};
node edge[maxm];
int head[maxn], cnt;
int dist[maxn], vis[maxn];
int n, m;
int sum, ed ;
void init(){
cnt = 0;
memset(head, -1, sizeof(head));
}
void add(int u, int v, int w){
edge[cnt] = {u, v, w, head[u]};
head[u] = cnt++;
edge[cnt] = {v, u, w, head[v]};
head[v] = cnt++;
}
void getmap(){
char str[5];
while(m--){
int a, b, c;
scanf("%d%d%d%s", &a, &b, &c, str);
add(a, b, c);
}
}
void bfs(int sx){
queueq;
for(int i = 1; i <= n; ++i){
vis[i] = 0;
dist[i] = 0;
}
vis[sx] = 1;
sum = 0;
ed = sx;
q.push(sx);
while(!q.empty()){
int u = q.front();
q.pop();
for(int i = head[u]; i != -1 ;i = edge[i].next){
int v = edge[i].v;
if(!vis[v]){
dist[v] = dist[u] + edge[i].w;
if(sum < dist[v]){
ed = v;
sum = dist[v];
}
vis[v] = 1;
q.push(v);
}
}
}
}
int main (){
while(scanf("%d%d", &n, &m) != EOF){
init();
getmap();
bfs(1);
bfs(ed);
printf("%d\n", sum);
}
return 0;
}