http://acm.hdu.edu.cn/showproblem.php?pid=3698 Let the light guide us Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 62768/32768 K (Java/Others) Total Submission(s): 759    Accepted Submission(s): 253 Problem Description Plain of despair was…

Let the light guide us Time Limit: 2000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 369864-bit integer IO format: %I64d      Java class name: Main   Plain of despair was once an ancient battlefield where those brave spiri…

Let the light guide us Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 62768/32768 K (Java/Others) Total Submission(s): 677    Accepted Submission(s): 226 Problem Description Plain of despair was once an ancient battlefield where those brave…

Description Plain of despair was once an ancient battlefield where those brave spirits had rested in peace for thousands of years. Actually no one dare step into this sacred land until the rumor that “there is a huge gold mine underneath the plain” s…

http://acm.hdu.edu.cn/showproblem.php?pid=3698 Let the light guide us Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 62768/32768 K (Java/Others) Total Submission(s): 821    Accepted Submission(s): 285 Problem Description Plain of despair was…

题目链接 and 题目大意 hdu3698 但是 hdu的数据比较弱,所以在这luogu提交吧UVA1490 Let the light guide us 有一个\(n*m\)的平原,要求每行选一个点,选\(n\)个点建造塔楼. 平原上每个点都有他自己的花费时间和魔法值. 为了正确控制塔楼,我们必须保证连续两排的每两座塔共用一个共同的魔法区域. 也就是要求每两行相邻的点都满足如下关系: 如果第\(i\)行选\(j\),第\(i+1\)行选\(k\),则需\(|j-k|≤f(i,j)+f(i+1,…

/* hdu 2857 Mirror and Light 计算几何 镜面反射 */ #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> using namespace std; const double DNF=100000001.0; int t; double x1,x2,y11,y2,xs,ys,xe,ye; struct Point { double x;…

Mirror and Light Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 650    Accepted Submission(s): 316 Problem Description The light travels in a straight line and always goes in the minimal path b…

题意: n个房间,每个有一定的钱,一个房间到另一个房间花费一定的时间,给你房间连接树,求在t时间内到达房间m能得到的最大钱数(从房间1(根)出发) 分析: 该题关键是最后要到达m,没有这个条件,就是基础的树形背包,哎,一开始没思路,放了一段时间,看看题解才明白,该题突破口,就是,你先想怎么判断不能到到达m的情况,自然想到最短路,对树先求一次最短路,在最短路上的点只能,过一次,其他得点过两次,把最短路上的点花费置零,剩下的就是基础的树形背包. #include <map> #include &l…

给出N*M矩阵.每一个点建立灯塔有花费.每一个点的灯塔有连接范围,求每一行都建立一个灯塔的最小花费,要求每相邻两行的灯塔能够互相连接.满足 |j-k|≤f(i,j)+f(i+1,k) DP思路,dp[i][j]=在第i行的j位置放置灯塔的最小花费.dp[i][j]=Min(dp[i-1][k]+a[i][j]),同一时候更新当前点能够覆盖该行的全部位置的最小值 要求上个区间的最小值,用线段树优化,否则超时 滚动数组,否则超内存 #include "stdio.h" #include &…