一開始我這么做超時(shí)了戴已，但是今天早上看了一下標(biāo)準(zhǔn)答案罗侯，感覺跟我的做法時(shí)間差不了多少啊，而且規(guī)定的時(shí)間是200ms音比，不可能超時(shí)啊妖滔，然后就檢查了一下隧哮，很驚喜的發(fā)現(xiàn)，我的方法居然過了座舍。
好開心啊
這個(gè)方法就是用X和Y換向來填充沮翔，但是速度肯定是比不上標(biāo)準(zhǔn)答案的那個(gè)啦，標(biāo)準(zhǔn)答案的那個(gè)簸州，一看就知道是很快的方法

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int input[100010];
bool cmp(int a, int b) { return a > b; }
int directionX[4] = { 1,0,-1,0 };
int directionY[4] = { 0,-1,0,1 };
int matrix[10000][10000];
int main() {
    int n; scanf("%d", &n);
    int rows, cols, half = ceil(sqrt(n)) ;
    for (int i = half; i <= n; i++)
        if (n % i == 0) {
            rows = i;
            cols = n / i;
            break;
        }
    for (int i = 0; i < n; i++) scanf("%d", &input[i]);
    sort(input, input + n, cmp);
    for (int i = 1; i <= cols; i++)matrix[1][i] = input[i - 1];
    int index = 0, index_X = 1, index_Y = cols , substract = 0;//用來指示方向,code用來表示第幾輪了
    for (int i = cols ; i < n;) {
        if (index % 2 == 0)substract++;
        int end;
        if (directionX[index]) end = rows - substract;
        else end = cols - substract;
        for (int j = 0; j < end; j++) {
            index_X += directionX[index];
            index_Y += directionY[index];
            matrix[index_X][index_Y] = input[i++];
        }
        index = (index + 1) % 4;
    }
    for (int i = 1; i <= rows; i++) {
        printf("%d", matrix[i][1]);
        for (int j = 2; j <= cols; j++)
            printf(" %d", matrix[i][j]);
        printf("\n");
    }
}

A1017 優(yōu)先隊(duì)列實(shí)現(xiàn)

#include<cstdio>
#include<vector>
#include<utility>
#include<queue>
#include<algorithm>
using namespace std;
int main() {
    priority_queue<pair<float, float>, vector<pair<float, float> >, greater<pair<float, float> > > customers;
    priority_queue<float,vector<float>,greater<float> > windows;
    int customers_num, windows_num,customers_sum = 0; scanf("%d %d", &customers_num, &windows_num);
    float close_time = 17 * 60;
    for (int i = 0; i < customers_num; i++) {
        float HH, SS, MM, proccesse; scanf("%f:%f:%f %f", &HH, &MM, &SS, &proccesse);
        MM += SS / 60 + HH * 60;
        if (proccesse > 60) proccesse = 60;
        if (MM <= close_time) { customers.push(pair<float, float>(MM, proccesse)); customers_sum++; }
    }
    float waitTimeSum = 0, open_time = 8 * 60;
    for (int i = 0; i < windows_num; i++)windows.push(open_time);
    while (customers.size() && windows.size()) {
        pair<float, float> now_customer = customers.top(); customers.pop();
        if (windows.top() > now_customer.first){waitTimeSum += windows.top() - now_customer.first;windows.push(windows.top() + now_customer.second);}
        else windows.push(now_customer.first + now_customer.second);
        windows.pop();
    }
    printf("%.1f\n", waitTimeSum / (float)customers_sum);
}

用隊(duì)列實(shí)現(xiàn)鉴竭，第一次做的時(shí)候溢出了
使用const int INF = 0x3fffffff 一定要注意它會(huì)不會(huì)有好幾個(gè)數(shù)字相加

#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int INF = 0x3fffffff;
int finish_time[1010], last_one[22];//用來記錄窗口的上一個(gè)處理的對(duì)象
int main(){
    queue<int> window[22];
    int windows_num, line_capacity,customers_num,queries,close_time = 17 * 60;
    scanf("%d %d %d %d",&windows_num,&line_capacity,&customers_num,&queries);
    for(int i = 0;i < windows_num;i++) last_one[i] = 480;
    int first_deal = min(customers_num,line_capacity * windows_num), index = 0;//先處理黃線中的顧客
    for(int i = 1;i <= first_deal;i++){
        int processe;scanf("%d",&processe);
        if(last_one[index] >= close_time) {last_one[index] = INF;finish_time[i] = INF;}
        else {processe += last_one[index];last_one[index] = processe;finish_time[i] = processe;}
        window[index++].push(i);
        if(index == windows_num) index = 0;//用條件判斷來代替除法
    }
    for(int i = first_deal + 1 ;i <= customers_num;i++){
        int processe;scanf("%d",&processe);
        int minindex = 0,min_num = INF;
        for(int j = 0;j < windows_num;j++)
            if(finish_time[window[j].front()] < min_num){//隊(duì)首元素最早的結(jié)束的歧譬，那么這個(gè)就會(huì)空出來岸浑，下一個(gè)元素就能進(jìn)來
                min_num = finish_time[window[j].front()];
                minindex = j;
            }
        window[minindex].pop();
        if(last_one[minindex] >= close_time) {last_one[minindex] = INF;finish_time[i] = INF;}
        else {processe += last_one[minindex];last_one[minindex] = processe;finish_time[i] = processe;}
        window[minindex].push(i);
        
    }
    for(int i = 0;i < queries;i++){
        int q;scanf("%d",&q);
        if(finish_time[q] == INF)printf("Sorry\n");
        else printf("%02d:%02d\n",finish_time[q] / 60,finish_time[q] % 60);
    }
}