cf gym101061J Cola(优先队列)

J. Cola

time limit per test
4 seconds
memory limit per test
64 megabytes

Problem Description
At Cola factory there is N bottles on a line. Also, there is a machine that pour cola in these bottles one by one. Every thing was working well. Suddenly, the machine started acting in a strange way !!
It started choosing a random bottle and pour a random quantity of cola in this bottle.
When a bottle is full, the rest of the cola goes to the bottle on the right of it, and if it was the last bottle, then the rest of the cola will be wasted .
As an engineer in this factory, you were given the capacity of every bottle in Litters. Also, you know the record of machine moves because it is stored in the Microcontroller memory of the machine. You are asked to compute the amount of wasted cola, and the amount of cola every bottle has at the end.

Input
Your program will be tested on one or more test cases. The first line of the input will be a single integer T the number of test cases. Followed by the test cases.
Every test case starts with two numbers N and M (1 ≤ N ≤ 105), (1 ≤ M ≤ 105) denoting the number of bottles, and the number of moves the machine had did.
Then follow N integers on a line separated by spaces. The integer Ci denotes the capacity of the ith bottle (1 ≤ Ci ≤ 109) where (1 ≤ i ≤ N).
M lines follow denoting the moves. Each line contains two integers X and Y (1 ≤ X ≤ N), (1 ≤ Y ≤ 109) means that the machine pouredY liters of cola in the Xth bottle .

Output
For each test case print two lines.
First line contains the amount of wasted cola.
The second line contains N integers separated by spaces denoting the amount of cola in every bottle at the end.

Example
input
2
5 3
5 4 3 2 1
3 4
4 4
1 2
6 3
3 4 5 5 6 7
3 3
1 8
5 9

output
2
2 0 3 2 1
0
3 4 4 0 6 3

工厂里的灌可乐的机器随机灌可乐，灌少了没事，多了漏到右边的瓶子里。右边没瓶子就漏出去浪费了。

比赛的时候没想到排序Orz，结果拜托学长用set做，完了之后和同学讨论说到排序，一拍脑袋，啊。。。。。。

把它的随机操作按照id从小到大排序，再写一个优先队列，每次操作之前先把所有id小于遍历到的操作的id的瓶子pop出去，一个瓶子灌满了也pop出去，降低时间复杂度。

#include <iostream>#include <string>#include <cstdio>#include <cmath>#include <cstring>#include <queue>#include <set>#include <algorithm>#define MM 100005#define mod 1000000007#define INF -0x3f3f3f3f#define PI acos(-1)#define ll long longusing namespace std;struct node{    ll id, liter, Max;    bool friend operator < (const node &a, const node &b)    {        return a.id > b.id;    }};struct pour{    ll ind, lit;    bool friend operator < (const pour &a, const pour &b)    {        return a.ind < b.ind;    }}arr[MM];priority_queue<node> Q;int ans[MM];int main(){    int T;    scanf("%d", &T);    while(T--)    {        int N, M;        scanf("%d %d", &N, &M);        node tmp;        for(int i = 1; i <= N; i++)        {            tmp.id = 1ll * i, tmp.liter = 0ll;            scanf("%I64d", &tmp.Max);            Q.push(tmp);        }        for(int i = 1; i <= M; i++)        {            scanf("%I64d %I64d", &arr[i].ind, &arr[i].lit);        }        sort(arr + 1, arr + M + 1);        ll cnt = 0ll;        node t;        for(int i = 1; i <= M; i++)        {            t = Q.top();            while(t.id < arr[i].ind)            {                ans[t.id] = t.liter;                Q.pop();                t = Q.top();            }            while(!Q.empty())            {                t = Q.top();                Q.pop();                if(t.Max - t.liter >= arr[i].lit)                {                    t.liter += arr[i].lit;                    arr[i].lit = 0;                    Q.push(t);                    break;                }                else                {                    ans[t.id] = t.Max;                    arr[i].lit -= t.Max - t.liter;                }            }            cnt += arr[i].lit;        }        while(!Q.empty())        {            t = Q.top();            ans[t.id] = t.liter;            Q.pop();        }        printf("%I64d\n", cnt);        for(int i = 1; i < N; i++)        {            printf("%d ", ans[i]);        }        printf("%d\n", ans[N]);    }    return 0;}

场下a过。
运行结果：