夏夜、

心若平似镜、何题不AC。

04 2014 档案

codeforces 420B Online Meeting
摘要：一道实现很蛋疼的题。必须静下理清思路，整理出各种情况。不然就会痛苦地陷入一大堆if..else里不能自拔。#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#i... 阅读全文

posted @ 2014-04-24 22:07 BMan、 阅读(393) 评论(0) 推荐(0)

POJ 3181 Dollar Dayz DP
摘要：f[i][j]=f[i-j][j]+f[i][j-1]，结果很大需要高精度。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#includ... 阅读全文

posted @ 2014-04-23 00:47 BMan、 阅读(164) 评论(0) 推荐(0)

POJ Ant Counting DP
摘要：dp[i][j]表示前i种蚂蚁组成元素个数为j的集合有多少种。则dp[i][j] = dp[i-1][j] + dp[i-1][j-1] + ... + dp[i-1][ max(0,j-a[i])];直接算的话复杂度为O(TA^2)状态的转移是一个区间内的数的和，所以再用一个数组f[i][j]记录... 阅读全文

posted @ 2014-04-22 16:01 BMan、 阅读(259) 评论(0) 推荐(0)

POJ 1742 Coins DP 01背包
摘要：dp[i][j]表示前i种硬币中取总价值为j时第i种硬币最多剩下多少个，-1表示无法到达该状态。a.当dp[i-1][j]>=0时，dp[i][j]=ci;b.当j-ai>=0&&dp[i-1][j-ai]>0时，dp[i][j]=dp[i-1][j-ai]-1;c.其他，dp[i][j]=-1So... 阅读全文

posted @ 2014-04-22 15:48 BMan、 阅读(140) 评论(0) 推荐(0)

POJ 1017 Packets 贪心 模拟
摘要：一步一步模拟，做这种题好累先放大的的，然后记录剩下的空位有多少，塞1*1和2*2的进去//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typede 阅读全文

posted @ 2014-04-07 19:54 BMan、 阅读(147) 评论(0) 推荐(0)

POJ 2394 Yogurt factory 贪心
摘要：维护前i个星期是哪个的花费最少（包括储存的费用） minv=min(minv,C_i)//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typede 阅读全文

posted @ 2014-04-07 17:15 BMan、 阅读(185) 评论(0) 推荐(0)

POJ 1328 Radar Installation 贪心
摘要：把每个点转成区间，排序，首先考虑最左边的区间[l1,r1]，要覆盖住这个点则需要在这个区间里的某个位置x放一个雷达，显然尽量放在靠右端是最好的，因为可以同时被右边的其他区间覆盖，但是要被右边的区间[l2,r2]覆盖还需满足l2#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pai 阅读全文

posted @ 2014-04-07 13:23 BMan、 阅读(136) 评论(0) 推荐(0)

POJ 2376 Cleaning Shifts 贪心
摘要：从左到右，如果需要覆盖当前点x那么就要选一个[l,r]满足l#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb(a) push(a)#define INF 0x1f1f1f1f#define lson idx T min(const T& 阅读全文

posted @ 2014-04-07 12:12 BMan、 阅读(142) 评论(0) 推荐(0)

2014浙大校赛
摘要：ZOJ 3767 Elevator求和，签到题//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define p 阅读全文

posted @ 2014-04-07 10:51 BMan、 阅读(296) 评论(0) 推荐(0)

POJ 3050 Hopscotch 暴力 DFS
摘要：DFS枚举每个点开始的全部长度为6的数字串，插入到set里面。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;# 阅读全文

posted @ 2014-04-06 11:58 BMan、 阅读(318) 评论(0) 推荐(0)

POJ 3187 Backward Digit Sums 枚举
摘要：用next_permutation枚举全排列。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define p 阅读全文

posted @ 2014-04-06 11:40 BMan、 阅读(143) 评论(0) 推荐(0)

POJ 2718 Smallest Difference 贪心构造
摘要：不用说，想让两个数的差最小，那这个两个数的位数要最接近。不妨设a>b;如果n是奇数 a就是最小的(n/2+1)位数，b就是最大的n/2位数如果是偶数 枚举每一对相邻的数，大的作为a的第一位，小的作为b的第一位，然后就一样了。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing 阅读全文

posted @ 2014-04-06 11:24 BMan、 阅读(169) 评论(0) 推荐(0)

POJ 3009 Curling 2.0 DFS
摘要：读懂题意就能写了，递归层数最多10，怎么写都能过。DFS：//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#de 阅读全文

posted @ 2014-04-06 10:38 BMan、 阅读(118) 评论(0) 推荐(0)

POJ 1979 Red and Black DFS
摘要：简单DFS。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb(a) push(a)#defi 阅读全文

posted @ 2014-04-05 23:32 BMan、 阅读(149) 评论(0) 推荐(0)

POJ 3169 Layout 差分约束系统
摘要：//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb(a) push(a)#define INF 阅读全文

posted @ 2014-04-05 22:55 BMan、 阅读(139) 评论(0) 推荐(0)

POJ 3723 Conscription 最小生成树
摘要：想清楚后能发现就要让我们求最小生成树（如果图连通的话，否则就是森林）//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pi 阅读全文

posted @ 2014-04-05 21:28 BMan、 阅读(167) 评论(0) 推荐(0)

POJ 3255 Roadblocks 次短路
摘要：和Dijksta求最短路一样，只是要维护两个数组：最短路d1,次短路d2。然后更新的时候注意细节。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull; 阅读全文

posted @ 2014-04-05 20:45 BMan、 阅读(174) 评论(0) 推荐(0)

UVA 11367 Full Tank? 最短路
摘要：以状态（u,fuel)为结点建图（把每个点拆成100个），表示在点u时还剩下fuel个单位的燃料，那么状态就可以这样转移：(u,fuel)->(u,fuel+i) ： 加i单位的燃料，所以这条边的权值就是 i*p[u];(u,fuel) ->(v,fuel-dist): 走到另一个点，其中dist为路径。当然直接这样做可能会超时，需要做两个小优化（做其中一个就能AC了）1. (u,fuel)->(u,fuel+i) 中的i只需为1就行了，这样就能边数从100^2降到100. 比如(u,fuel)->(u,fuel+3)可以用 (u,fuel)->(u,fuel+1 阅读全文

posted @ 2014-04-05 19:31 BMan、 阅读(207) 评论(0) 推荐(0)

UVA 10269 Adventure of Super Mario 最短路
摘要：（i,j)表示当前在结点j，还剩下i次使用魔法的机会。以(i,j)为结点建图，求最短路。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typed 阅读全文

posted @ 2014-04-05 13:30 BMan、 阅读(159) 评论(0) 推荐(0)

UVA 10603 Fill 最短路
摘要：//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb(a) push(a)#define INF 阅读全文

posted @ 2014-04-05 13:28 BMan、 阅读(163) 评论(0) 推荐(0)

POJ 2431 Expedition 优先队列
摘要：由于油箱是无限的，所以每次经过加油站都不选择加油，等到后面油不够了，再来“反悔”，把前面经过的油站的油加上，尽可能选量大的,用优先队列维护。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef un 阅读全文

posted @ 2014-04-04 17:01 BMan、 阅读(159) 评论(0) 推荐(0)

POJ 3253 Fence Repair 贪心
摘要：每次合并选最短的两块木板，用优先队列优化。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb( 阅读全文

posted @ 2014-04-04 16:02 BMan、 阅读(139) 评论(0) 推荐(0)

POJ 3609 Saruman's Army 贪心
摘要：从坐开始，贪心选择每个点标记还是不标记。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb(a 阅读全文

posted @ 2014-04-04 15:50 BMan、 阅读(277) 评论(0) 推荐(0)

POJ 3617 Best Cow Line 贪心
摘要：不能单纯比较头尾两个字符，应该比较头尾两个字符串。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define 阅读全文

posted @ 2014-04-04 15:33 BMan、 阅读(146) 评论(0) 推荐(0)

POJ 2386 Lake Counting DFS入门题
摘要：//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair pii;#define pb(a) push(a)#define INF 阅读全文

posted @ 2014-04-04 15:23 BMan、 阅读(159) 评论(0) 推荐(0)

POJ 1852 Ants 思维
摘要：假设每只蚂蚁都是一样的，那么相遇后调转方向和相遇后擦肩而过继续走是一样的。//#pragma comment(linker, "/STACK:1024000000,1024000000")#include#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair 阅读全文

posted @ 2014-04-04 15:14 BMan、 阅读(207) 评论(0) 推荐(0)

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