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2012年8月19日

POJ 1274 The Perfect Stall(二分图匹配)

摘要: The Perfect StallTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 14447Accepted: 6612DescriptionFarmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering problems, all the stalls in the new barn are different. For th 阅读全文

posted @ 2012-08-19 00:51 kuangbin 阅读(1638) 评论(0) 推荐(0) 编辑

POJ 3020 Antenna Placement(二分图匹配)

摘要: Antenna PlacementTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 4739Accepted: 2359DescriptionThe Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile phone nets in Sweden. The most striking reason why they got the job, is their discovery of a 阅读全文

posted @ 2012-08-19 00:41 kuangbin 阅读(494) 评论(0) 推荐(0) 编辑

2012年8月18日

POJ 3041 Asteroids(二分匹配模板题)

摘要: AsteroidsTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 10288Accepted: 5556DescriptionBessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K asteroids (1 <= K <= 10,000), which are convenientl 阅读全文

posted @ 2012-08-18 23:30 kuangbin 阅读(544) 评论(0) 推荐(0) 编辑

POJ 1511 Invitation Cards(最短路)

摘要: Invitation CardsTime Limit: 8000MSMemory Limit: 262144KTotal Submissions: 14556Accepted: 4710DescriptionIn the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. T 阅读全文

posted @ 2012-08-18 20:57 kuangbin 阅读(460) 评论(0) 推荐(0) 编辑

POJ 1850 Code(排列数组)

摘要: CodeTime Limit: 1000MSMemory Limit: 30000KTotal Submissions: 6093Accepted: 2865DescriptionTransmitting and memorizing information is a task that requires different coding systems for the best use of the available space. A well known system is that one where a number is associated to a character sequ 阅读全文

posted @ 2012-08-18 20:54 kuangbin 阅读(315) 评论(0) 推荐(0) 编辑

POJ 2240 Arbitrage(求正环)

摘要: ArbitrageTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 11008Accepted: 4629DescriptionArbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British po 阅读全文

posted @ 2012-08-18 19:39 kuangbin 阅读(572) 评论(0) 推荐(0) 编辑

POJ 1125 Stockbroker Grapevine(任意两点的最短路,FLoyed 算法)

摘要: Stockbroker GrapevineTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 21124Accepted: 11474DescriptionStockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the stockbrokers to give your employer the tactical edge in th 阅读全文

posted @ 2012-08-18 18:50 kuangbin 阅读(401) 评论(0) 推荐(0) 编辑

POJ 2253 Frogger(最短路的变形)

摘要: FroggerTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 17842Accepted: 5818DescriptionFreddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunsc 阅读全文

posted @ 2012-08-18 17:13 kuangbin 阅读(761) 评论(0) 推荐(0) 编辑

大数乘加小数模板

摘要: //大数乘小数void mult(char c[],int m,char t[]){ char s[100]; int len=strlen(c); for(int i=0;i<len;i++) s[len-i-1]=c[i]-'0'; int flag,add=0; for(int i=0;i<len;i++) { int k=s[i]*m+add; if(k>=10) { s[i]=k%10; add=k/10; flag=1; ... 阅读全文

posted @ 2012-08-18 09:01 kuangbin 阅读(993) 评论(0) 推荐(0) 编辑

POJ 1062 昂贵的聘礼(最短路Dijkstra)

摘要: 昂贵的聘礼Time Limit: 1000MSMemory Limit: 10000KTotal Submissions: 27365Accepted: 7639Description年轻的探险家来到了一个印第安部落里。在那里他和酋长的女儿相爱了,于是便向酋长去求亲。酋长要他用10000个金币作为聘礼才答应把女儿嫁给他。探险家拿不出这么多金币,便请求酋长降低要求。酋长说:"嗯,如果你能够替我弄到大祭司的皮袄,我可以只要8000金币。如果你能够弄来他的水晶球,那么只要5000金币就行了。"探险家就跑到大祭司那里,向他要求皮袄或水晶球,大祭司要他用金币来换,或者替他弄来其他的东 阅读全文

posted @ 2012-08-18 01:45 kuangbin 阅读(605) 评论(0) 推荐(0) 编辑

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