MooFest
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 5697 | Accepted: 2481 |
Description
Every year, Farmer John's N (1 <= N <= 20,000) cows attend "MooFest",a social gathering of cows from around the world. MooFest involves a variety of events including haybale stacking, fence jumping, pin the tail on the farmer, and of course, mooing. When the cows all stand in line for a particular event, they moo so loudly that the roar is practically deafening. After participating in this event year after year, some of the cows have in fact lost a bit of their hearing.
Each cow i has an associated "hearing" threshold v(i) (in the range 1..20,000). If a cow moos to cow i, she must use a volume of at least v(i) times the distance between the two cows in order to be heard by cow i. If two cows i and j wish to converse, they must speak at a volume level equal to the distance between them times max(v(i),v(j)).
Suppose each of the N cows is standing in a straight line (each cow at some unique x coordinate in the range 1..20,000), and every pair of cows is carrying on a conversation using the smallest possible volume.
Compute the sum of all the volumes produced by all N(N-1)/2 pairs of mooing cows.
Each cow i has an associated "hearing" threshold v(i) (in the range 1..20,000). If a cow moos to cow i, she must use a volume of at least v(i) times the distance between the two cows in order to be heard by cow i. If two cows i and j wish to converse, they must speak at a volume level equal to the distance between them times max(v(i),v(j)).
Suppose each of the N cows is standing in a straight line (each cow at some unique x coordinate in the range 1..20,000), and every pair of cows is carrying on a conversation using the smallest possible volume.
Compute the sum of all the volumes produced by all N(N-1)/2 pairs of mooing cows.
Input
* Line 1: A single integer, N
* Lines 2..N+1: Two integers: the volume threshold and x coordinate for a cow. Line 2 represents the first cow; line 3 represents the second cow; and so on. No two cows will stand at the same location.
* Lines 2..N+1: Two integers: the volume threshold and x coordinate for a cow. Line 2 represents the first cow; line 3 represents the second cow; and so on. No two cows will stand at the same location.
Output
* Line 1: A single line with a single integer that is the sum of all the volumes of the conversing cows.
Sample Input
4 3 1 2 5 2 6 4 3
Sample Output
57
Source
读题真是个大问题啊！！！一开始读了好久也没读懂，还是看的别人题解才明白了题意。
好吧，题意好歹明白了，可是问题又来了，怎么办？
归根到底就是一个动态改值并求和的问题，树状数组（线段树当然也可）。
我发现我的树状数组真不会用，就是用的不熟练，以前做的都直接套，我都没弄懂啥意思，一次锻炼吧。
往往成功就差那么一步啊，确是咫尺天涯。
#include <cstdio> #include <iostream> #include <sstream> #include <cmath> #include <cstring> #include <cstdlib> #include <string> #include <vector> #include <map> #include <set> #include <queue> #include <stack> #include <algorithm> using namespace std; #define ll long long #define _cle(m, a) memset(m, a, sizeof(m)) #define repu(i, a, b) for(int i = a; i < b; i++) #define MAXN 20005 struct P{ ll v, p; bool operator < (const P& t) const { return t.v > v; } }cow[MAXN]; ll c_ount[MAXN] = {0}; ll total[MAXN] = {0}; ll sum_tot[MAXN] = {0}; int n; ll lowbit(ll x) { return x & (-x); } void add(int x, int d, ll c[]) { while(x < MAXN) { c[x] += d; x += lowbit(x); } } ll Sum(ll x, ll c[]) { ll ret = 0; while(x > 0) { ret += c[x]; x -= lowbit(x); } return ret; } int main() { scanf("%d", &n); repu(i, 1, n + 1) scanf("%lld%lld", &cow[i].v, &cow[i].p); sort(cow + 1, cow + n + 1); repu(i, 1, n + 1) sum_tot[i] = sum_tot[i - 1] + cow[i].p; ll sum = 0, num_cow = 0, sum_total = 0; add(cow[1].p, 1, c_ount); add(cow[1].p, cow[1].p, total); repu(i, 2, n + 1) { num_cow = Sum(cow[i].p, c_ount); sum_total = Sum(cow[i].p, total); sum += cow[i].v * (num_cow * cow[i].p - sum_total + (sum_tot[i - 1] - sum_total - (i - 1 - num_cow) * cow[i].p)); add(cow[i].p, 1, c_ount); add(cow[i].p, cow[i].p, total); } printf("%lld ", sum); return 0; }