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?
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
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
【题意】给你一棵树，每个节点有一个权值，M次询问，给出u,v,a,b,求权值在区间[a,b]中的和。
【分析】数据实在是太水了，暴力都能过，这里说一下正确的树剖，离线往线段树插值得写法。将每个询问分成两部分，[1,b]-[1,a-1],放进一个 集合排序，从小到大取出，将节点权值同样 排序，若当前权值<=询问中的权值，则插入线段树。
#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?' ':' '); } return 0; }