Ch’s gift
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1354 Accepted Submission(s): 496
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
Source
分析：由于没有修改操作，一个显然的想法是离线处理所有问题
将询问拆成1-x，1-y，1-LCA(x,y)，则处理的问题转化为从根到节点的链上的问题。
解决这个问题，我们可以在dfs时向treap插入当前的数，在退出时删除这个数，并且每次维护在该点上的答案。
当然也可以将所有的查询和点权排序，用树链剖分做这个题，在线段树上面插入就ok。
将询问拆成1-x，1-y，1-LCA(x,y)，则处理的问题转化为从根到节点的链上的问题。
解决这个问题，我们可以在dfs时向treap插入当前的数，在退出时删除这个数，并且每次维护在该点上的答案。
当然也可以将所有的查询和点权排序，用树链剖分做这个题，在线段树上面插入就ok。
下面给出AC代码：
1 #include<stdio.h> 2 #include<vector> 3 #include<string.h> 4 #include<algorithm> 5 using namespace std; 6 #define LL long long 7 vector<LL> G[100005]; 8 typedef struct 9 { 10 LL id; 11 LL x, y; 12 LL l, r; 13 }Quer; 14 Quer s[100005], cnt[100005]; 15 LL tre[445005]; 16 LL n, val[100005], son[100005], fa[100005], siz[100005], dep[100005]; 17 LL k, top[100005], rak[100005], id[100005], anl[100005], anr[100005]; 18 bool comp1(Quer a, Quer b) 19 { 20 if(a.l<b.l) 21 return 1; 22 return 0; 23 } 24 bool compr(Quer a, Quer b) 25 { 26 if(a.r<b.r) 27 return 1; 28 return 0; 29 } 30 void Sechs(LL u, LL p) 31 { 32 LL v, i; 33 fa[u] = p; 34 dep[u] = dep[p]+1; 35 for(i=0;i<G[u].size();i++) 36 { 37 v = G[u][i]; 38 if(v==p) 39 continue; 40 Sechs(v, u); 41 siz[u] += siz[v]+1; 42 if(son[u]==0 || siz[v]>siz[son[u]]) 43 son[u] = v; 44 } 45 } 46 void Sechr(LL u, LL p) 47 { 48 LL v, i; 49 top[u] = p; 50 rak[u] = ++k, id[k] = u; 51 if(son[u]==0) 52 return; 53 Sechr(son[u], p); 54 for(i=0;i<G[u].size();i++) 55 { 56 v = G[u][i]; 57 if(v==son[u] || v==fa[u]) 58 continue; 59 Sechr(v, v); 60 } 61 } 62 63 void Update(LL l, LL r, LL x, LL a, LL b) 64 { 65 LL m; 66 if(l==r) 67 { 68 tre[x] += b; 69 return; 70 } 71 m = (l+r)/2; 72 if(a<=m) 73 Update(l, m, x*2, a, b); 74 else 75 Update(m+1, r, x*2+1, a, b); 76 tre[x] = tre[x*2]+tre[x*2+1]; 77 } 78 LL Query(LL l, LL r, LL x, LL a, LL b) 79 { 80 LL m, sum = 0; 81 if(l>=a && r<=b) 82 return tre[x]; 83 m = (l+r)/2; 84 if(a<=m) 85 sum += Query(l, m, x*2, a, b); 86 if(b>=m+1) 87 sum += Query(m+1, r, x*2+1, a, b); 88 return sum; 89 } 90 LL TreQuery(LL x, LL y) 91 { 92 LL sum, p1, p2; 93 p1 = top[x], p2 = top[y], sum = 0; 94 while(p1!=p2) 95 { 96 if(dep[p1]<dep[p2]) 97 swap(p1, p2), swap(x, y); 98 sum += Query(1, n, 1, rak[p1], rak[x]); 99 x = fa[p1], p1 = top[x]; 100 } 101 if(dep[x]>dep[y]) 102 swap(x, y); 103 sum += Query(1, n, 1, rak[x], rak[y]); 104 return sum; 105 } 106 107 int main(void) 108 { 109 LL i, j, x, y, q; 110 while(scanf("%lld%lld", &n, &q)!=EOF) 111 { 112 for(i=1;i<=n;i++) 113 G[i].clear(); 114 for(i=1;i<=n;i++) 115 { 116 scanf("%lld", &val[i]); 117 cnt[i].id = i; 118 cnt[i].r = val[i]; 119 } 120 sort(cnt+1, cnt+n+1, compr); 121 for(i=1;i<=n-1;i++) 122 { 123 scanf("%lld%lld", &x, &y); 124 G[x].push_back(y); 125 G[y].push_back(x); 126 } 127 memset(siz, 0, sizeof(siz)); 128 memset(son, 0, sizeof(son)); 129 k = 0; 130 Sechs(1, 0); 131 Sechr(1, 1); 132 for(i=1;i<=q;i++) 133 { 134 scanf("%lld%lld%lld%lld", &s[i].x, &s[i].y, &s[i].l, &s[i].r); 135 s[i].id = i; 136 } 137 138 sort(s+1, s+q+1, comp1); 139 memset(tre, 0, sizeof(tre)); 140 j = 1; 141 for(i=1;i<=q;i++) 142 { 143 while(cnt[j].r<s[i].l && j<=n) 144 { 145 Update(1, n, 1, rak[cnt[j].id], cnt[j].r); 146 j += 1; 147 } 148 anl[s[i].id] = TreQuery(s[i].x, s[i].y); 149 } 150 151 sort(s+1, s+q+1, compr); 152 memset(tre, 0, sizeof(tre)); 153 j = 1; 154 for(i=1;i<=q;i++) 155 { 156 while(cnt[j].r<=s[i].r && j<=n) 157 { 158 Update(1, n, 1, rak[cnt[j].id], cnt[j].r); 159 j += 1; 160 } 161 anr[s[i].id] = TreQuery(s[i].x, s[i].y); 162 } 163 164 printf("%lld", anr[1]-anl[1]); 165 for(i=2;i<=q;i++) 166 printf(" %lld", anr[i]-anl[i]); 167 printf("\n"); 168 } 169 return 0; 170 }
