POJ 3162
『题目链接』POJ 3162
『题目类型』bit区间查询最值+树形DP
✡Problem：
一棵n个节点的树。wc爱跑步，跑n天，第i天从第i个节点开始跑步，每次跑到距第i个节点最远的那个节点（产生了n个距离），现在要在这n个距离里取连续的若干天，使得这些天里最大距离和最小距离的差小于M，问怎么取使得天数最多？
✡Answer：
参考：http://blog.csdn.net/angon823/article/details/52314522
这题和HDU 2196有相同的部分，先求一下树上每个点到其它点的最远距离。问题就转化为：求一个最长的符合要求的区间，即区间里 最大值 - 最小值 < M。
我用的是树状数组初始化和查询最值，当然也可以用线段树；可以求区间最值了，那么我们用尺取法就可以求出最长的区间了：维护两个指针i，j；让i=j=1；先右移j，直到不符合情况时再右移i，之后每次更新最大长度就好了。
fread输入挂能省1s
我看了看数据，1e6，就想弄弄输入挂试试，结果如下，上面的是输入挂。fread真的牛逼，但一定不要忘了要在main函数中加2句：fread(Buf, 1, BUF, stdin); fwrite(Out, 1, ou - Out, stdout);
如果不加第一句，输入时就会rt；如果不加第二句，就没有输出。
『时间复杂度』\(O(nlogn)\)
✡Code：
#include <map>
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <set>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
#define cle(a,v) memset(a,(v),sizeof(a))
#define fo(i,a,b) for(int i=(a);i<=(b);i++)
#define fd(i,a,b) for(int i=(a);i>=(b);i--)
#define ll long long
const int maxn = 1e6 + 7;
int n, m, u, v, w, tot, head[maxn], dp[maxn][3], num[maxn];
struct Edge {
int v, w, next;
} edges[maxn << 1];
void added(int u, int v, int w) {
edges[tot] = Edge{v, w, head[u]};
head[u] = tot++;
}
void init() {
tot = 0;
cle(head, -1);
}
void dfsson(int u, int fa) {
for (int i = head[u]; ~i; i = edges[i].next) {
int v = edges[i].v;
if (v == fa) continue;
dfsson(v, u);
int temporary = dp[v][0] + edges[i].w;
if (temporary >= dp[u][0]) {
dp[u][1] = dp[u][0];
dp[u][0] = temporary;
}
else if (temporary > dp[u][1]) {
dp[u][1] = temporary;
}
}
}
void dfsfa(int u, int fa) {
for (int i = head[u]; ~i; i = edges[i].next) {
int v = edges[i].v;
if (v == fa) continue;
if (dp[u][0] == dp[v][0] + edges[i].w) {
dp[v][2] = max(dp[u][2], dp[u][1]) + edges[i].w;
}
else {
dp[v][2] = max(dp[u][2], dp[u][0]) + edges[i].w;
}
dfsfa(v, u);
}
}
#define lowb(x) (x&-x)
struct node {
int ma, mi;
} bit[maxn];
void Binit() {
for (int i = 1; i <= n; i++) {
bit[i].ma = num[i];
bit[i].mi = num[i];
for (int j = 1; j < lowb(i); j <<= 1) {
bit[i].ma = max(bit[i].ma, bit[i - j].ma);
bit[i].mi = min(bit[i].mi, bit[i - j].mi);
}
}
}
node Bquery(int l, int r) {
node ans = node{num[r], num[r]};
while (true) {
ans.ma = max(ans.ma, num[r]);
ans.mi = min(ans.mi, num[r]);
if (r == l) break;
for (r--; r - l >= lowb(r); r -= lowb(r)) {
ans.ma = max(ans.ma, bit[r].ma);
ans.mi = min(ans.mi, bit[r].mi);
}
}
return ans;
}
int main() {
freopen("1.in", "r", stdin);
init();
scanf("%d%d", &n, &m);
for (int i = 2; i <= n; i++) {
scanf("%d%d", &v, &w);
added(i, v, w); added(v, i, w);
}
dfsson(1, -1);
dfsfa(1, -1);
for (int i = 1; i <= n; i++)
num[i] = max(dp[i][0], dp[i][2]);
// fo(i, 1, n) {
// printf("%d ", num[i]);
// } printf("\n");
Binit();
int len = 1, i = 1, j = 1;
while (j <= n) {
node a = Bquery(i, j);
while (a.ma - a.mi < m && j <= n) {
j++;
a = Bquery(i, j);
}
len = max(len, j - i);
i++;
}
printf("%d\n", len);
return 0;
}
✡Code_fread()：
#include <map>
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <set>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
#define cle(a,v) memset(a,(v),sizeof(a))
#define fo(i,a,b) for(int i=(a);i<=(b);i++)
#define fd(i,a,b) for(int i=(a);i>=(b);i--)
#define ll long long
const int maxn = 1e6 + 7;
int n, m, u, v, w, tot, head[maxn], dp[maxn][3], num[maxn];
struct Edge {
int v, w, next;
} edges[maxn << 1];
void added(int u, int v, int w) {
edges[tot] = Edge{v, w, head[u]};
head[u] = tot++;
}
void init() {
tot = 0;
cle(head, -1);
}
void dfsson(int u, int fa) {
for (int i = head[u]; ~i; i = edges[i].next) {
int v = edges[i].v;
if (v == fa) continue;
dfsson(v, u);
int temporary = dp[v][0] + edges[i].w;
if (temporary >= dp[u][0]) {
dp[u][1] = dp[u][0];
dp[u][0] = temporary;
}
else if (temporary > dp[u][1]) {
dp[u][1] = temporary;
}
}
}
void dfsfa(int u, int fa) {
for (int i = head[u]; ~i; i = edges[i].next) {
int v = edges[i].v;
if (v == fa) continue;
if (dp[u][0] == dp[v][0] + edges[i].w) {
dp[v][2] = max(dp[u][2], dp[u][1]) + edges[i].w;
}
else {
dp[v][2] = max(dp[u][2], dp[u][0]) + edges[i].w;
}
dfsfa(v, u);
}
}
#define lowb(x) (x&-x)
struct node {
int ma, mi;
} bit[maxn];
void Binit() {
for (int i = 1; i <= n; i++) {
bit[i].ma = num[i];
bit[i].mi = num[i];
for (int j = 1; j < lowb(i); j <<= 1) {
bit[i].ma = max(bit[i].ma, bit[i - j].ma);
bit[i].mi = min(bit[i].mi, bit[i - j].mi);
}
}
}
node Bquery(int l, int r) {
node ans = node{num[r], num[r]};
while (true) {
ans.ma = max(ans.ma, num[r]);
ans.mi = min(ans.mi, num[r]);
if (r == l) break;
for (r--; r - l >= lowb(r); r -= lowb(r)) {
ans.ma = max(ans.ma, bit[r].ma);
ans.mi = min(ans.mi, bit[r].mi);
}
}
return ans;
}
const int BUF = 40000000;
char Buf[BUF], *buf = Buf;
const int OUT = 20000000;
char Out[OUT], *ou = Out; int Outn[30], Outcnt;
inline void write(int x) {
if (!x)*ou++ = 48;
else {
for (Outcnt = 0; x; x /= 10)Outn[++Outcnt] = x % 10 + 48;
while (Outcnt)*ou++ = Outn[Outcnt--];
}
}
inline void writell(ll x) {
if (!x)*ou++ = 48;
else {
for (Outcnt = 0; x; x /= 10)Outn[++Outcnt] = x % 10 + 48;
while (Outcnt)*ou++ = Outn[Outcnt--];
}
}
inline void writechar(char x) {*ou++ = x;}
inline void writeln() {*ou++ = '\n';}
inline void read(int&a) {for (a = 0; *buf < 48; buf++); while (*buf > 47)a = a * 10 + *buf++ -48;}
int main() {
freopen("1.in", "r", stdin);
init();
//DON'T FOGET THIS!!!
fread(Buf, 1, BUF, stdin);
read(n), read(m);
for (int i = 2; i <= n; i++) {
read(v), read(w);
added(i, v, w); added(v, i, w);
}
dfsson(1, -1);
dfsfa(1, -1);
for (int i = 1; i <= n; i++)
num[i] = max(dp[i][0], dp[i][2]);
// fo(i, 1, n) {
// printf("%d ", num[i]);
// } printf("\n");
Binit();
int len = 1, i = 1, j = 1;
while (j <= n) {
node a = Bquery(i, j);
while (a.ma - a.mi < m && j <= n) {
j++;
a = Bquery(i, j);
}
len = max(len, j - i);
i++;
}
write(len);
writeln();
//DON'T FOGET THIS!!!
fwrite(Out, 1, ou - Out, stdout);
return 0;
}