HDU5861-Road

Road

                                                                    Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                              Total Submission(s): 1287    Accepted Submission(s): 363

Problem Description

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 ai to village bi, and you need to guarantee that the segments between ai and bi 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 wi(1<=wi<=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).

 

Output

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

 

Sample Input

4 31 2 31 33 42 4

 

Sample Output

355

 

Author

BUPT

 

Source

2016 Multi-University Training Contest 10

 

Recommend

wange2014

 

题意：n个点将一条线段分成n-1份，点的编号从左往右1~n，每条路你只能打开一次，打开后每天都收费，当然打开后你也可以选择关上，但是关上后这条路就再也不能打开了。每条线段有一个cost值，有q次询问，每次询问给出两个点U,V，表示从U到V，你必须保证U->V上的线段都打开才能过去，问在保证能过去的情况下，每天的最小花费

解题思路：线段树，预处理出每段路的打开时间和关闭时间即可

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, m;LL w[200009], sum[200009];int a[200009], b[200009];int l[200009], r[200009];int x[200009 << 2];void update(int k, int l, int r, int ll, int rr, int val){if (l >= ll&&r <= rr) { x[k] = val; return; }int mid = (l + r) >> 1;if (x[k] != -1) x[k << 1] = x[k << 1 | 1] = x[k], x[k] = -1;if (ll <= mid) update(k << 1, l, mid, ll, rr, val);if (rr>mid) update(k << 1 | 1, mid + 1, r, ll, rr, val);}int query(int k, int l, int r,int p){if (l == r || x[k] != -1) return x[k];int mid = (l + r) >> 1;if(mid>=p) return query(k << 1, l, mid, p);else return query(k << 1 | 1, mid + 1, r, p);}int main(){while (~scanf("%d%d", &n, &m)){for (int i = 1; i<n; i++) scanf("%lld", &w[i]);memset(sum, 0, sizeof sum);memset(a, -1, sizeof a);memset(b, -1, sizeof b);memset(x, -1, sizeof x);for (int i = 1; i <= m; i++){scanf("%d%d", &l[i], &r[i]);if (l[i]>r[i]) swap(l[i], r[i]);}for (int i = m; i >= 1; i--) update(1, 1, n-1, l[i], r[i] - 1, i);for (int i = 1; i <= n; i++){int k=query(1, 1, n - 1, i);if (k != -1) a[i] = k;}memset(x, -1, sizeof x);for (int i = 1; i <= m; i++) update(1, 1, n - 1, l[i], r[i] - 1, i);for (int i = 1; i <= n; i++){int k = query(1, 1, n - 1, i);if (k != -1) b[i] = k + 1;}for (int i = 1; i<n; i++){if (a[i] == -1) continue;sum[a[i]] += w[i];sum[b[i]] -= w[i];}for (int i = 1; i <= m; i++){sum[i] += sum[i - 1];printf("%lld\n", sum[i]);}}return 0;}