Processor(二分搜索,优先队列)

题意:

给出n个任务完成区间以及工作量,求出最快处理速度.

思路:

用二分搜索匹配最优速度,其中还用到优先队列辅助.

枚举所有截止时间,在此时间内的都进队,按照工作截止时间越早越先出队,如果截止时间不及枚举的截止时间,则错过

代码:

#include<cstdio>#include<queue>#include<iostream>#include<string>#include<algorithm>using namespace std;const int N = 10005;int n;struct Work {double r, d, w;bool operator < (const Work& a)const {return d>a.d;//工作按照结束时间在队内调整}}work[N];bool cmp(Work a, Work b) {return a.r < b.r;}void init() {scanf("%d", &n);for(int i=0; i<n; i++) scanf("%lf%lf%lf", &work[i].r, &work[i].d, &work[i].w);sort(work, work+n, cmp);//工作按照起始时间进队}bool judge(int m) {priority_queue<Work> qu;int wn = 0;for(int i=1; i<=20000; i++) {int sum = m;while(work[wn].r<i && wn<n) qu.push(work[wn++]);while(!qu.empty() && sum) {//每秒内工作量没用完而且还有工作没完成.Work tmp = qu.top();qu.pop();if(tmp.d < i) return false;//作业已错过if(sum >= tmp.w) sum -= tmp.w;else {tmp.w -= sum;qu.push(tmp);sum = 0;}}if(qu.empty() && wn == n) return true;}return false;}int deal() {int l=0, r=5000, mid;while(l<=r) {mid = (l+r)/2;if(judge(mid)) r = mid-1;else l = mid+1;}return l;}int main() {int kase;scanf("%d", &kase);while(kase--) {init();printf("%d\n", deal());}return 0;}