9-3（线段树维护最大最小值）

HDU 5861 Road

There are n villages along a high way, and divided the high way into n-1 segments. Each segment would charge a certain amount of money for being open for one day, and you can open or close an arbitrary segment in an arbitrary day, but you can open or close the segment for just one time, because the workers would be angry if you told them to work multiple period.

We know the transport plan in the next m days, each day there is one cargo need to transport from village aiai to village bibi, and you need to guarantee that the segments between aiai and bibi are open in the i-th day. Your boss wants to minimize the total cost of the next m days, and you need to tell him the charge for each day.

(At the beginning, all the segments are closed.)

Input

Multiple test case. For each test case, begins with two integers n, m(1<=n,m<=200000), next line contains n-1 integers. The i-th integer wiwi(1<=wiwi<=1000) indicates the charge for the segment between village i and village i+1 being open for one day. Next m lines, each line contains two integers ai,bi(1≤ai,bi<=n,ai!=bi)ai,bi(1≤ai,bi<=n,ai!=bi).

Output

For each test case, output m lines, each line contains the charge for the i-th day.

Sample Input

4 3
1 2 3
1 3
3 4
2 4

Sample Output

3
5
5

#include <bits/stdc++.h>
using namespace std;
int n, m;
const int maxn = 200010;
const int INF = 0x3f3f3f3f;
int cost[maxn];
vector<int> vec_s[maxn], vec_t[maxn];
struct Tree{
    int l, r, ddl;
    int open;
}tree[maxn*5];

void build(int node){
    tree[node].ddl = 0;
    tree[node].open = INF;
    if(tree[node].l == tree[node].r) return ;
    int mid = (tree[node].l + tree[node].r)/2;
    tree[node*2].l = tree[node].l;
    tree[node*2].r = mid;
    tree[node*2+1].l = mid+1;
    tree[node*2+1].r = tree[node].r;
    build(node*2);
    build(node*2+1);
}

void pushdown(int node){
    tree[node*2].ddl = max(tree[node*2].ddl, tree[node].ddl);
    tree[node*2+1].ddl = max(tree[node*2+1].ddl, tree[node].ddl);
    tree[node*2].open = min(tree[node*2].open, tree[node].open);
    tree[node*2+1].open = min(tree[node*2+1].open, tree[node].open);
}

void update(int node, int s, int e, int c){
    if(s <= tree[node].l && tree[node].r <= e){
        if(tree[node].l != tree[node].r) pushdown(node);
        tree[node].ddl = max(tree[node].ddl, c);
        tree[node].open = min(tree[node].open, c);
        return;
    }
    pushdown(node);
    int mid = (tree[node].l + tree[node].r) / 2;
    if(s <= mid)
        update(node*2, s, e, c);
    if(e > mid)
        update(node*2+1, s, e, c);

    tree[node].open = max(tree[node*2].open, tree[node*2+1].open);
    tree[node].ddl = min(tree[node*2].ddl, tree[node*2+1].ddl);
    return;
}

void query(int node, int &tar, int& s, int& t){
    if(tree[node].l == tree[node].r && tree[node].l == tar){
        s = tree[node].open;
        t = tree[node].ddl;
        if(s == INF) s = t = 0;
        return;
    }
    pushdown(node);
    int mid = (tree[node].l + tree[node].r)/2;
    if(tar <= mid)
        query(node*2, tar, s, t);
    else 
        query(node*2+1, tar, s, t);
}

int main()
{
    while(cin >> n >> m){
        for(int i = 1; i < n; i++){
            scanf("%d", &cost[i]);
        }

        tree[1].l = 1, tree[1].r = n-1;
        build(1);
        for(int i = 1; i <= m; i++){
            vec_s[i].clear(), vec_t[i].clear();
            int s, e;
            scanf("%d%d", &s, &e);
            if(s>e) swap(s, e);   //题目没说这里的s 一定小于 e 啊！！！！！！
            update(1, s, e-1, i);
        }
        for(int i = 1; i <= n-1; i++){
            int s = 0, t = 0;
            query(1, i, s, t);
            vec_s[s].push_back(i);
            vec_t[t].push_back(i);
        }
        int sum = 0;
        for(int i = 1; i <= m; i++){
            for(int j = 0; j < vec_s[i].size(); j++){
                sum += cost[vec_s[i][j]];
            }
            printf("%d\n", sum);
            for(int j = 0; j < vec_t[i].size(); j++)
                sum -= cost[vec_t[i][j]];
        }
    }
    return 0;
}