BZOJ 1509: [NOI2003]逃学的小孩( 树形dp )

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 ", 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
[Submit][Status][Discuss]

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

HINT

Source