树形dp求出某个点的最长3条链a,b,c(a>=b>=c), 然后以这个点为交点的最优解一定是a+2b+c.好像还有一种做法是求出树的直径然后乱搞...
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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
using namespace std;
typedef long long ll;
const int maxn = 200009;
int N;
ll X[maxn], Y[maxn], Z[maxn], F[maxn], ans;
struct edge {
int to, w;
edge* next;
} E[maxn << 1], *pt = E, *head[maxn];
void AddEdge(int u, int v, int w) {
pt->to = v;
pt->w = w;
pt->next = head[u];
head[u] = pt++;
}
inline int read() {
char c = getchar();
int ret = 0;
for(; !isdigit(c); c = getchar());
for(; isdigit(c); c = getchar()) ret = ret * 10 + c - ‘0‘;
return ret;
}
void Init() {
N = read();
int m = read();
while(m--) {
int u = read() - 1, v = read() - 1, w = read();
AddEdge(u, v, w);
AddEdge(v, u, w);
}
}
void DFS0(int x, int fa = -1) {
X[x] = Y[x] = 0;
for(edge* e = head[x]; e; e = e->next) if(e->to != fa) {
DFS0(e->to, x);
Z[x] = max(Z[x], X[e->to] + e->w);
if(Z[x] > Y[x]) swap(Z[x], Y[x]);
if(Y[x] > X[x]) swap(X[x], Y[x]);
}
}
void DFS1(int x, int fa = -1) {
for(edge* e = head[x]; e; e = e->next) if(e->to != fa) {
F[e->to] = F[x] + e->w;
if(X[e->to] + e->w == X[x])
F[e->to] = max(F[e->to], Y[x] + e->w);
else
F[e->to] = max(F[e->to], X[x] + e->w);
DFS1(e->to, x);
}
}
inline void upd(ll &x, ll &y, ll &z) {
if(y > x) swap(x, y);
if(z > x) swap(x, z);
if(z > y) swap(y, z);
ans = max(x + (y << 1) + z, ans);
}
int main() {
Init();
DFS0(0);
DFS1(F[0] = 0);
ans = 0;
for(int i = 0; i < N; i++)
F[i] <= Z[i] ? upd(X[i], Y[i], Z[i]) : upd(X[i], Y[i], F[i]);
printf("%lld\n", ans);
return 0;
}
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1509: [NOI2003]逃学的小孩
Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 528 Solved: 285
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Description
Input
第一行是两个整数N（3 ? N ? 200000）和M，分别表示居住点总数和街道总数。以下M行，每行给出一条街道的信息。第i+1行包含整数Ui、Vi、Ti（1?Ui, Vi ? N，1 ? Ti ? 1000000000），表示街道i连接居住点Ui和Vi，并且经过街道i需花费Ti分钟。街道信息不会重复给出。
Output
仅包含整数T，即最坏情况下Chris的父母需要花费T分钟才能找到Chris。
Sample Input
4 3
1 2 1
2 3 1
3 4 1
Sample Output
4