给出一棵树 ，和边的权值
求权值最长的一条直径
两次bfs求
第一次以任意点开始 BFS求出第一个端点
第二次以第一次得到的端点 BFS求出第二个端点
#include #include #include #include #include #include #include #include #include #include #include using namespace std; #include #include #include #include #include #include typedef long long LL; typedef long double LD; #define pi acos(-1.0) #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 typedef pair PI; typedef pair PP; #ifdef _WIN32 #define LLD "%I64d" #else #define LLD "%lld" #endif const int MAXN = 200100; const int INF = 999999; //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;} //inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;} //inline void print(LL x){printf(LLD, x);puts("");} //inline void read(int &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}} struct Edge { int to,next,val; } edge[MAXN*2]; int head[MAXN],d[MAXN],tol; bool vis[MAXN]; void init() { tol=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,int w) { edge[tol].to=v,edge[tol].val=w,edge[tol].next=head[u]; head[u]=tol++; edge[tol].to=u,edge[tol].val=w,edge[tol].next=head[v]; head[v]=tol++; } int bfs(int u) { int point,big=0; memset(vis,false,sizeof(vis)); queueq; q.push(u); vis[u]=true; d[u]=0; while(!q.empty()) { u=q.front(); q.pop(); for(int i=head[u]; ~i; i=edge[i].next) { int v=edge[i].to; if(!vis[v]) { d[v]=d[u]+edge[i].val; if(d[v]>big) { big=d[v]; point=v; } vis[v]=true; q.push(v); } } } return point; } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); #endif int t,a,b,c,n,m; while(scanf("%d%d",&n,&m)!=EOF) { init(); for(int i=0; i
转载于:https://www.cnblogs.com/kewowlo/p/4088333.html
