Doing Homework again 点我挑战题目 题意分析 给出n组数据,每组数据中有每份作业的deadline和score,如果不能按期完成,则要扣相应score,求每组数据最少扣除的score是多少. 典型的贪心策略. 既然是要求最少的扣分,那么肯定是要先完成分数最多的.所以可以推出按照分数排序.那么最佳策略应该是在deadline当天完成作业,如果那天已经占用,只能在deadline-1天完成,如果那天也被占用了,就只能在deadline-2天完成--直到推到第1天,如果还被占用的话…

Doing Homework again Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8638    Accepted Submission(s): 5090 Problem Description Ignatius has just come back school from the 30th ACM/ICPC. Now he ha…

Problem Description Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will r…

//大意理解 先排序 最早交的里面选最大值 扫描完了加没写的 排序后 应该是早交的和扣分多的在前 用结构体吧/*#include<stdio.h>#include<stdio.h>int cmp(void const* a,void const*b){ if(*(st*)a->t==*(st*)b->t) return *(st*)b->kou-*(st*)b->kou; else return *(st*)a->t-*(st*)b->t;}ty…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784    Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…

Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994    Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…

Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color…

卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…

Rectangles    HDOJ(2056) http://acm.hdu.edu.cn/showproblem.php?pid=2056 题目描述:给2条线段,分别构成2个矩形,求2个矩形相交面积. 算法:先用快速排斥判断2个矩形是否相交.若不相交,面积为0.若相交,将x坐标排序去中间2个值之差,y坐标也一样.最后将2个差相乘得到最后结果. 这题是我大一的时候做过的,当时一看觉得很水,写起来发现其实没我想的那么水.分了好几类情况没做出来.今天看了点关于判断线段相交的知识,想起了这题便拿来练…