POJ 3162 - Walking Race(树形DP)

题目：

http://poj.org/problem?id=3162

题意：

树中节点的直径，最大值与最小值的差值不超过m的最大区间。

思路：

步骤1：求树的直径。

步骤2：求出区间，维护两个单调队列，一个递增一个递减。

AC.

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e6+5;int n, m;int tol, head[maxn];struct Edge{    int to, w, next;}edge[maxn*2];void addedge(int u, int v, int w){    edge[tol].to = v;    edge[tol].w = w;    edge[tol].next = head[u];    head[u] = tol++;}int dis[maxn], dson[maxn], pre[maxn];int d[maxn];void dfs(int u, int fa){    int tmp = 0;    for(int i = head[u]; ~i; i = edge[i].next) {        int v = edge[i].to, w = edge[i].w;        if(v == fa) continue;        dfs(v, u);        dson[u] = max(dson[u], dson[v] + w);    }}void dfs1(int u, int fa){    int max1 = 0, max2 = 0, tmp;    int v1;    int v, w;    for(int i = head[u]; ~i; i = edge[i].next) {        v = edge[i].to; w = edge[i].w;        if(v == fa) continue;        tmp = dson[v] + w;        if(tmp > max1) {            max2 = max1; max1 = tmp;            v1 = v;        }        else if(tmp == max1 || tmp > max2) {            max2 = tmp;        }    }    if(u != 1) {        tmp = dis[u];        v = -1;        if(tmp > max1) {            max2 = max1; max1 = tmp;            v1 = v;        }        else if(max1 == tmp || tmp > max2) max2 = tmp;    }    for(int i = head[u]; ~i; i = edge[i].next) {        v = edge[i].to; w = edge[i].w;        if(v == fa) continue;        if(v == v1) dis[v] = max2+w; // printf("v= %d: %d\n", v, dis[v]);        else dis[v] = max1+w; // printf("**v= %d: %d\n", v, dis[v]);        dfs1(v, u);    }}int qmax[maxn], qmin[maxn];void solve(){    int ans = 1;    int s = 1,t = 1;    int t1=1, t2=1, i1=1, i2=1;    qmax[t1++] = d[1]; qmin[t2++] = d[1];    while(t < n) {        if(qmax[i1] - qmin[i2] <= m) {            t++;            int tmp = d[t];            while(i1 < t1 && qmax[t1-1] < tmp) t1--;            qmax[t1++] = tmp;            while(i2 < t2 && qmin[t2-1] > tmp) t2--;            qmin[t2++] = tmp;        }        else {            int tmp = d[s++];            if(qmax[i1] == tmp) i1++;            if(qmin[i2] == tmp) i2++;        }        if(qmax[i1] - qmin[i2] <= m) {            //printf("%d %d: ", qmax[i1], qmin[i2]);            //printf("(%d, %d)\n", s, t);            ans = max(ans, t-s+1);        }    }    printf("%d\n", ans);}void init(){    tol = 0;    memset(head, -1, sizeof(head));    memset(dis, 0, sizeof(dis));    memset(d, 0, sizeof(d));    memset(dson, 0, sizeof(dson));    memset(qmax, 0, sizeof(qmax));    memset(qmin, 0, sizeof(qmin));}int main(){    //freopen("inn", "r", stdin);    while(~scanf("%d%d", &n, &m)) {        init();        for(int u = 2; u <= n; ++u) {            int v, w;            scanf("%d%d", &v, &w);            addedge(u, v, w);            addedge(v, u, w);        }        dfs(1, -1);        dfs1(1, -1);        for(int i = 1; i <= n; ++i) {            //printf("%d %d: ", dson[i], dis[i]);            d[i] = max(dson[i], dis[i]);           // printf("%d\n", d[i]);        }        solve();    }    return 0;}