【POJ】【3308】Paratroopers】的更多相关文章

Word Puzzles Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 10782 Accepted: 4076 Special Judge Description Word puzzles are usually simple and very entertaining for all ages. They are so entertaining that Pizza-Hut company started using t…
不可以理解的是,测评站上的0ms是怎么搞出来的. 这一题在建立超级源点和超级汇点后就变得温和可爱了.其实它本身就温和可爱.对比了能够找到的题解: (1)艾德蒙·卡普算法(2)迪尼克算法(3)改进版艾德蒙·卡普算法(MY METHOD) 不去管那个0ms的吧,那么(3)号算法最为美妙[它的别名是:ISAP],时间可观. 这个就算是一个ISAP的模板吧(除了输入的难看的几行外,其余均是标准的大米饼牌模板!) #include<stdio.h> #include<algorithm> #…
Time Limit: 3000MSMemory Limit: 65536K Total Submissions: 27109Accepted: 7527 Description David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water…
Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 13849Accepted: 4851 Description In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to be . Given your test scores an…
Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 19026Accepted: 8466 Description The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributo…
1.POJ1258 水水的prim……不过poj上硬是没过,wikioi上的原题却过了 #include<cstring> #include<algorithm> #include<cstdio> using namespace std; ,inf=1e8; ],g[maxn+][maxn+],ans=,n; ]; int main() { scanf("%d",&n); ;i<=n;++i) ;j<=n;++j) { scanf…
流星雨撞击地球(平面直角坐标第一象限),问到达安全地带的最少时间. 对于每颗流星雨i,在ti时刻撞击(xi,yi)点,同时导致(xi,yi)和上下左右相邻的点在ti以后的时刻(包括t)不能再经过(被封锁).安全地带为永远不会被封锁的点. 简单bfs,开始WA在把平面空间上限当成300*300,但根据题目,这只是有流星雨撞击的范围.实际可走的空间理论上没上限,但分析可得,离原点最近的安全地带一定在(302,302)范围内,所以应可把数组至少开为303*303. 后来WA在把G[0][0]==1的情…
题目大意:给n个数,一个长度为k(k<n)的闭区间从0滑动到n,求滑动中区间的最大值序列和最小值序列. 最大值和最小值是类似的,在此以最大值为例分析. 数据结构要求:能保存最多k个元素,快速取得最大值,更新时删去“过期”元素和“不再有希望”的元素,安放新元素. 单调队列的基本概念百度百科讲得比较清楚了:http://baike.baidu.com/view/3771451.htm 我的大致思路是: 1. 每个元素存储为结构体,包含它的秩和值.维护最大长度为k的单调队列,保证所有元素的秩都在区间内…
Time Limit: 3000MSMemory Limit: 65536K Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication…
2295: [POJ Challenge]我爱你啊 Time Limit: 1 Sec  Memory Limit: 128 MB Description ftiasch是个十分受女生欢迎的同学,所以她总是收到许多情书.虽然她十分有魅力,然而她却是个低调的人.因此她从来不会告诉别人她到底收到了多少情书. ftiasch的好朋友1tthinking想知道她到底收到了多少情书.1tthinking知道,ftiasch每次收到一封情书,就会在日记最后写下一个包含"luvletter"子序列的…