push panel hose either 插值 dfs dir sample start

Ch’s gift

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 662 Accepted Submission(s): 229

Problem Description Mr. Cui is working off-campus and he misses his girl friend very much. After a whole night tossing and turning, he decides to get to his girl friend‘s city and of course, with well-chosen gifts. He knows neither too low the price could a gift be since his girl friend won‘t like it, nor too high of it since he might consider not worth to do. So he will only buy gifts whose price is between [a,b].
There are n cities in the country and (n-1) bi-directional roads. Each city can be reached from any other city. In the ith city, there is a specialty of price ci Cui could buy as a gift. Cui buy at most 1 gift in a city. Cui starts his trip from city s and his girl friend is in city t. As mentioned above, Cui is so hurry that he will choose the quickest way to his girl friend(in other words, he won‘t pass a city twice) and of course, buy as many as gifts as possible. Now he wants to know, how much money does he need to prepare for all the gifts?

Input There are multiple cases.

For each case:
The first line contains tow integers n,m(1≤n,m≤10^5), representing the number of cities and the number of situations.
The second line contains n integers c1,c2,...,cn(1≤ci≤10^9), indicating the price of city i‘s specialty.
Then n-1 lines follows. Each line has two integers x,y(1≤x,y≤n), meaning there is road between city x and city y.
Next m line follows. In each line there are four integers s,t,a,b(1≤s,t≤n;1≤a≤b≤10^9), which indicates start city, end city, lower bound of the price, upper bound of the price, respectively, as the exact meaning mentioned in the description above

Output Output m space-separated integers in one line, and the ith number should be the answer to the ith situation.

Sample Input 5 3 1 2 1 3 2 1 2 2 4 3 1 2 5 4 5 1 3 1 1 1 1 3 5 2 3

Sample Output 7 1 4

【題意】給你一棵樹，每個節點有一個權值，M次詢問，給出u,v,a,b,求權值在區間[a,b]中的和。

【分析】數據實在是太水了，暴力都能過，這裏說一下正確的樹剖，離線往線段樹插值得寫法。將每個詢問分成兩部分，[1,b]-[1,a-1],放進一個 集合排序，從小到大取出，將節點權值同樣 排序，若當前權值<=詢問中的權值，則插入線段樹。

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define lson(x) ((x<<1))
#define rson(x) ((x<<1)+1)
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 1e5+50;;
const int M = 17;
const int mod = 2520;
const int mo=123;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
int dep[N],siz[N],fa[N],id[N],son[N],val[N],top[N];
int num,n,m;
ll ans[N];
vector<int>edg[N];
vector<pii>vec;
struct query{
    int u,v,x,id;
    bool operator <(const query &d)const{
        return x<d.x;
    }
}q[N*2];
void dfs1(int u, int f, int d) {
    dep[u] = d;
    siz[u] = 1;
    son[u] = 0;
    fa[u] = f;
    for (int v : edg[u]) {
        if (v == f) continue;
        dfs1(v, u, d + 1);
        siz[u] += siz[v];
        if (siz[son[u]] < siz[v])
            son[u] = v;
    }
}
void dfs2(int u, int tp) {
    top[u] = tp;
    id[u] = ++num;
    if (son[u]) dfs2(son[u], tp);
    for (int v : edg[u]) {
        if (v == fa[u] || v == son[u]) continue;
        dfs2(v, v);
    }
}

struct Tree {
    int l,r;
    ll sum;
};
Tree tree[4*N];
void pushup(int x) {
    tree[x].sum=tree[lson(x)].sum+tree[rson(x)].sum;
}
void build(int l,int r,int v) {
    tree[v].l=l;
    tree[v].r=r;
    if(l==r) {
        tree[v].sum=0;
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,v*2);
    build(mid+1,r,v*2+1);
    pushup(v);
}
void update(int o,int v,int val) {
    if(tree[o].l==tree[o].r) {
        tree[o].sum= val;
        return ;
    }
    int mid = (tree[o].l+tree[o].r)/2;
    if(v<=mid)
        update(o*2,v,val);
    else
        update(o*2+1,v,val);
    pushup(o);
}
ll querySum(int x,int l,int r) {
    if (tree[x].l >= l && tree[x].r <= r) {
        return tree[x].sum;
    }
    int mid = (tree[x].l + tree[x].r) / 2;
    ll ans = 0;
    if (l <= mid) ans += querySum(lson(x),l,r);
    if (r > mid) ans += querySum(rson(x),l,r);
    return ans;
}
ll Qsum(int u,int v) {
    int tp1 = top[u], tp2 = top[v];
    ll ans = 0;
    while (tp1 != tp2) {
        if (dep[tp1] < dep[tp2]) {
            swap(tp1, tp2);
            swap(u, v);
        }
        ans +=querySum(1,id[tp1], id[u]);
        u = fa[tp1];
        tp1 = top[u];
    }
    if (dep[u] > dep[v])swap(u, v);
    ans +=querySum(1,id[u], id[v]);
    return ans;
}
void init(){
    for(int i=0;i<N;i++)edg[i].clear();
    met(tree,0);met(son,0);vec.clear();met(ans,0);
}
int main() {
    while(~scanf("%d%d",&n,&m)) {
        init();
        for(int i=1,x; i<=n; i++)scanf("%d",&x),vec.pb(mp(x,i));
        for(int i=1,u,v; i<n; i++) {
            scanf("%d%d",&u,&v);
            edg[u].pb(v);edg[v].pb(u);
        }
        num = 0;
        dfs1(1,0,1);
        dfs2(1,1);
        sort(vec.begin(),vec.end());
        for(int i=1;i<=m;i++){
            int x,y,a,b;
            scanf("%d%d%d%d",&x,&y,&a,&b);
            q[i]=query{x,y,a-1,-i};
            q[i+m]=query{x,y,b,i};
        }
        sort(q+1,q+2*m+1);
        int now=0;
        build(1,num,1);
        for(int i=1;i<=2*m;i++){
            while(now<n&&vec[now].first<=q[i].x){
                update(1,id[vec[now].second],vec[now].first);
                now++;
            }
            ans[abs(q[i].id)]+=Qsum(q[i].u,q[i].v)*(q[i].id>0?1:-1);
        }
        for(int i=1;i<=m;i++)printf("%lld%c",ans[i],i==m?‘\n‘:‘ ‘);
    }
    return 0;
}

HDU 6162 Ch’s gift （樹剖 + 離線線段樹）