标签：other stdout ble etc 小问题 names get lse name
题目链接
题意，求一棵树被所有直径经过的边的条数。
这题是我们8.25KS图论的最后一题，当时我果断打了暴力求所有直径然后树上差分统计的方法，好像有点小问题，boom0了。
考完改这题，改了好久，各种各样的小bug，至少有七八个。。。
思路：先随便找一条直径，然后从一个端点开始遍历这条直径，如果当前点能分叉出一条直径，那么这条直径后面的点都不可能被所有直径穿过了，于是我们找到第一个能分叉的点，然后再从这个点往回遍历，找到第一个能分叉的点，这2个点中间的路径的边即为所求。
代码就算了吧，打这题时我就没想过要有可读性
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
#define INF 2147483647
#define Open(s) freopen(s".in","r",stdin);freopen(s".out","w",stdout);
#define Close fclose(stdin); fclose(stdout);
const int MAXN = 200010;
inline int read(){
int s = 0, w = 1;
char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') w = -1; ch = getchar(); }
while(ch >= '0' && ch <= '9') { s = s * 10 + ch - '0'; ch = getchar(); }
return s * w;
}
struct Edge{
int next, to, dis;
}e[MAXN << 1];
int num, head[MAXN], flag[MAXN];
inline void Add(int from, int to, int dis){
e[++num] = (Edge){ head[from], to, dis };
head[from] = num;
}
int pre[MAXN], in[MAXN];
int n, ans, Maxu, ans1;
long long Max;
void dfs(int u, int fa, long long dep){
if(dep > Max) Max = dep, Maxu = u;
for(int i = head[u]; i; i = e[i].next)
if(e[i].to != fa)
dfs(e[i].to, u, dep + e[i].dis);
}
void DFS(int u, int fa, long long dep){
pre[u] = fa;
if(dep > Max) Max = dep, Maxu = u;
for(int i = head[u]; i; i = e[i].next)
if(e[i].to != fa)
DFS(e[i].to, u, dep + e[i].dis);
}
bool existOther(int u, int fa, long long dep){
if(dep == Max) return true;
for(int i = head[u]; i; i = e[i].next)
if(e[i].to != fa)
if(existOther(e[i].to, u, dep + e[i].dis)) return true;
return false;
}
int A, B, C;
int main(){
Open("diameter");
n = read();
for(int i = 1; i < n; ++i){
A = read(); B = read(); C = read();
Add(A, B, C); Add(B, A, C);
}
Max = -1; dfs(1, 0, 0);
Max = -1; DFS(Maxu, 0, 0);
int now = Maxu, o = 0, fa = 0, Plus = 0;
long long deep = 0;
while(now) flag[now] = 1, now = pre[now];
for(int i = 1; i <= n; ++i) if(!flag[i]) pre[i] = 0;
now = Maxu;
while(now){
Plus = 0;
for(int i = head[now]; i; i = e[i].next)
if(e[i].to != pre[now] && e[i].to != fa){
if(existOther(e[i].to, now, deep + e[i].dis)){
o = now;
break;
}
}
else if(e[i].to == pre[now]) Plus = e[i].dis;
if(!o && !pre[now]) o = now;
if(o) break;
fa = now;
now = pre[now];
deep += Plus;
} //找到第一个分叉的点
ans = 0; //往回遍历
now = o;
fa = pre[now];
deep = Max - deep;
while(now != Maxu){
int nxt;
++ans;
for(int i = head[now]; i; i = e[i].next)
if(pre[e[i].to] != now && e[i].to != fa){
if(existOther(e[i].to, now, deep + e[i].dis)){
printf("%I64d\n%d\n", Max, ans - 1);
//system("pause");
return 0;
}
}
else if(pre[e[i].to] == now) nxt = e[i].to, Plus = e[i].dis;
fa = now;
now = nxt;
deep += Plus;
}
printf("%I64d\n%d\n", Max, ans); //如果都不能分叉
Close;
//system("pause");
return 0;
}
标签：other stdout ble etc 小问题 names get lse name
原文地址：https://www.cnblogs.com/Qihoo360/p/9541297.html
