UVA 1422 （LA 4254）- Processor

An ``early adopter" Mr. Kim bought one of the latest notebooks which has a speed-controlled processor. The processor is able to operate at variable speed. But the higher the speed, the higher the power consumption is. So, to execute a set of programs, adjusting the speed of the processor dynamically results in energy-efficient schedules. We are concerned in a schedule to minimize the maximum speed of the processor.

The processor shall execute a set of programs and each program P_i is given having a starting time r_i, a deadline d_i, and workw_i. When the processor executes the programs, for each programP_i, the work w_i should be done on the processor within the interval[r_i, d_i] to completeP_i. Also, the processor does not have to execute a program in a contiguous interval, that is, it can interrupt the currently running program and later resume it at the interrupted point. It is assumed thatr_i, d_i , andw_i are given positive integers. Recall that the processor can execute the programs at variable speed. If the processor runs the programP_i with work w_i at a constant speeds,then it takes time to completeP_i. We also assume that the available speeds are positive integers, that is, the processor operates only at integer points of speed. The speed is unbounded and the processor may operate at sufficiently large speeds to complete all the programs. The processor should complete all the given programs and the goal is to find a schedule minimizing the maximum of the speeds at which the processor operates.

For example, there are five programs P_i with the interval[r_i, d_i] and workw_i, i = 1,..., 5 , where[r₁, d₁] = [1, 4],[r₂, d₂] = [3, 6],[r₃, d₃] = [4, 5],[r₄, d₄] = [4, 7],[r₅, d₅] = [5, 8] andw₁ = 2, w₂ = 3,w₃ = 2, w₄ = 2,w₅ = 1. Then the Figure 1 represents a schedule which minimizes the maximum speed at which the processor operates. The maximum speed is 2 in this example.

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Input 

Your program is to read from standard input. The input consists of T test cases. The number of test cases T(1T20) is given on the first line of the input. The first line of each test case contains an integern (1n10, 000), the number of given programs which the processor shall execute. In the nextn lines of each test case, the i-th line contain three integer numbers r_i,d_i, and w_i, representing the starting time, the deadline, and the work of the programP_i, respectively, where 1r_i <d_i20, 000,1w_i1, 000.

Output 

Your program is to write to standard output. Print exactly one line for each test case. The line contains the maximum speed of a schedule minimizing the maximum speed at which the processor operates to complete all the given programs.

Sample Input 

3 5 1 4 2 3 6 3 4 5 2 4 7 2 5 8 1 6 1 7 25 4 8 10 7 10 5 8 11 5 10 13 10 11 13 5 8 15 18 10 20 24 16 8 15 33 11 14 14 1 6 16 16 19 12 3 5 12 22 25 10

Sample Output 

2 5 7

题意：有n个任务，每个任务有3个参数 ri，di和wi，表示必须在[ri, di]内执行，工作量为wi，处理器速度可变，当速度为s时，工作量为w的工作要执行s/w个单位时间。每个任务可以分成多块在不连续时间段执行。求处理器最大速度最小值。

二分很容易想到，但是判断不好办。

先二分答案，然后就是判断处理器速度为s是能否执行完所有工作。根据贪心策略：枚举每一单位时间，对于任意一单位时间，优先选择di（di即上面的区间[ri, di]中的di）小的任务。当然选择的任务开始时间必须在此之前，原因不多说，肯定选择最紧迫的事情做。可以用优先队列实现。

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <string>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <set>using namespace std;#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#endif#define S64I(a) scanf(iform, &a)#define P64I(a) printf(oform, a)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const int maxn = 10000 + 20;struct Mission {int r, d, w;bool operator < (const Mission & b) const {return d > b.d;}};Mission ms[maxn];int n;int cmp(Mission a, Mission b) {return a.r < b.r;}int check(int p) {priority_queue<Mission> Q;int msn = 0;for(int i=1; i<=20000; i++) {while(msn < n && ms[msn].r < i) {Q.push(ms[msn++]);}int time = p;while(time && !Q.empty()) {Mission t = Q.top();Q.pop();if(t.d < i) return 0;if(time >= t.w) {time -= t.w;} else {t.w -= time;time = 0;Q.push(t);}}if(msn == n && Q.empty()) return 1;}return 0;}int main() {int T;scanf("%d", &T);while(T--) {scanf("%d", &n);int sum = 0;for(int i=0 ; i<n; i++) {scanf("%d%d%d", &ms[i].r, &ms[i].d, &ms[i].w);sum += ms[i].w;}sort(ms, ms+n, cmp);int l = 0, r = sum;while(l < r) {int mid = l + (r - l) / 2;if(check(mid)) r = mid;else l = mid + 1;}printf("%d\n", l);}return 0;}