04 2015 档案
摘要:bestcoder#37_1001 字符串,dfsRikka with string Accepts: 395 Submissions: 2281 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/...
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摘要:codeforces#297div2_d bfs,2*2法判断矩阵里的矩形D. Arthur and Wallstime limit per test2 secondsmemory limit per test512 megabytesinputstandard inputoutputstandar...
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摘要:codeforces#297div2_c 贪心C. Ilya and Stickstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputIn the ...
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摘要:codeforces#297div2_b 贪心,字符串Pasha got a very beautiful stringsfor his birthday, the string consists of lowercase Latin letters. The letters in the stri...
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摘要:poj2983——差分约束,bellman_fordIs the Information Reliable?Time Limit:3000MSMemory Limit:131072KTotal Submissions:11560Accepted:3658DescriptionThe galaxy w...
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摘要:poj1201——差分约束,spfaIntervalsTime Limit:2000MSMemory Limit:65536KTotal Submissions:22553Accepted:8530DescriptionYou are given n closed, integer interval...
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摘要:图的邻接表存储struct Edge{ int v; ll w; Edge *next;};Edge e[maxn*10];void add_edge(int u,int v,ll w) ///插入邻接表的首部而非尾部,避免遍历{ Edge *pre=&e[u]; Ed...
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摘要:hiho1093 ——spfa时间限制:10000ms单点时限:1000ms内存限制:256MB描述万圣节的晚上,小Hi和小Ho在吃过晚饭之后,来到了一个巨大的鬼屋!鬼屋中一共有N个地点,分别编号为1..N,这N个地点之间互相有一些道路连通,两个地点之间可能有多条道路连通,但是并不存在一条两端都是同...
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摘要:poj1716——差分约束Integer IntervalsTime Limit:1000MSMemory Limit:10000KTotal Submissions:13029Accepted:5516DescriptionAn integer interval [a,b], a =2,0=2,s...
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摘要:poj1113——凸包WallTime Limit:1000MSMemory Limit:10000KTotal Submissions:31132Accepted:10497DescriptionOnce upon a time there was a greedy King who ordere...
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摘要:求凸包—— graham_scan算法先按Y-X排序,在按对p0的极角排序,然后进行扫瞄bool cmp(Point A,Point B){ //return (A-vert[0])*(B-vert[0])>EPS; double tmp=(A-vert[0])*(B-vert[0]);...
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摘要:poj2187 求平面最远点对,garham_scan算法求凸包Beauty ContestTime Limit:3000MSMemory Limit:65536KTotal Submissions:29666Accepted:9180DescriptionBessie, Farmer John's...
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摘要:bestcoder#36 1002 哈希GunnerTime Limit: 8000/4000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 726Accepted Submission(s...
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摘要:poj1584——判断凸包,判断点是否在多边形内A Round Peg in a Ground HoleTime Limit:1000MSMemory Limit:10000KTotal Submissions:5441Accepted:1729DescriptionThe DIY Furnitur...
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摘要:poj1408——叉积求多边形面积FishnetTime Limit:1000MSMemory Limit:10000KTotal Submissions:1853Accepted:1185DescriptionA fisherman named Etadokah awoke in a very s...
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摘要:重载Point其中减法表示向量,乘法表示叉积struct Point{ double x,y; friend double operator*(const Point A,const Point B) { return A.x*B.y-A.y*B.x; } ...
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摘要:求直线与线段的交点求交点定比分点法,用叉积求面积,用面积比代替线段长度比Point inter_point(Point A,Point B,Point C,Point D)///返回直线AB和线段CD的交点{ if(!intersect(A,B,C,D)) return {-INF*1.0,0...
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摘要:poj1039——计算几何 求直线与线段交点,判断两条直线是否相交PipeTime Limit:1000MSMemory Limit:10000KTotal Submissions:9439Accepted:2854DescriptionThe GX Light Pipeline Company s...
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