https://blog.youkuaiyun.com/Wall_Fzh-cnhttps://blog.youkuaiyun.com/https://blog.youkuaiyun.com/Wall_F/article/details/11717099https://blog.youkuaiyun.com/Wall_F/article/details/11717099Wall_FMon, 16 Sep 2013 00:59:55 +0800i连wi的流量，费用为0的边。 i+n->t，连wi的流量，费用为0的边。  城市与城市之间，i>j+n 连流量为1，费用为cost的边。跑最小费用最大流即可。 最后判断最]]>https://blog.youkuaiyun.com/Wall_F/article/details/11664709https://blog.youkuaiyun.com/Wall_F/article/details/11664709Wall_FSat, 14 Sep 2013 00:05:08 +0800https://blog.youkuaiyun.com/Wall_F/article/details/11265137https://blog.youkuaiyun.com/Wall_F/article/details/11265137Wall_FFri, 06 Sep 2013 20:29:35 +0800https://blog.youkuaiyun.com/Wall_F/article/details/11152169https://blog.youkuaiyun.com/Wall_F/article/details/11152169Wall_FThu, 05 Sep 2013 16:28:58 +0800https://blog.youkuaiyun.com/Wall_F/article/details/11019765https://blog.youkuaiyun.com/Wall_F/article/details/11019765Wall_FTue, 03 Sep 2013 21:12:43 +0800https://blog.youkuaiyun.com/Wall_F/article/details/10948337https://blog.youkuaiyun.com/Wall_F/article/details/10948337Wall_FMon, 02 Sep 2013 20:52:34 +0800https://blog.youkuaiyun.com/Wall_F/article/details/10020691https://blog.youkuaiyun.com/Wall_F/article/details/10020691Wall_FSat, 17 Aug 2013 11:01:40 +0800https://blog.youkuaiyun.com/Wall_F/article/details/10012195https://blog.youkuaiyun.com/Wall_F/article/details/10012195Wall_FFri, 16 Aug 2013 22:42:32 +0800https://blog.youkuaiyun.com/Wall_F/article/details/9991073https://blog.youkuaiyun.com/Wall_F/article/details/9991073Wall_FThu, 15 Aug 2013 21:49:16 +0800T流代表路径，在最短路树上的点连边既可，容量为1. 似乎题目保证有解。 #include #include #include #include #include #include #include #include #include #include using namespace std; const int maxn = 1610; const int INF =]]>https://blog.youkuaiyun.com/Wall_F/article/details/9978763https://blog.youkuaiyun.com/Wall_F/article/details/9978763Wall_FThu, 15 Aug 2013 10:15:27 +0800b->a->c->d呢？当然可以，但是由于拆边之后没有负费用，我们肯定能找到一条更优的路径。比如说a->c->d。由于是最小费用，所以通过这样的方法就保证了不会走]]>https://blog.youkuaiyun.com/Wall_F/article/details/9965631https://blog.youkuaiyun.com/Wall_F/article/details/9965631Wall_FWed, 14 Aug 2013 15:39:39 +0800https://blog.youkuaiyun.com/Wall_F/article/details/9954883https://blog.youkuaiyun.com/Wall_F/article/details/9954883Wall_FTue, 13 Aug 2013 20:52:23 +0800https://blog.youkuaiyun.com/Wall_F/article/details/9927167https://blog.youkuaiyun.com/Wall_F/article/details/9927167Wall_FMon, 12 Aug 2013 15:47:39 +0800B之间通过所有路径一小时之内能通过最大的车辆(Maxflow)/所有边上通过最大车流量（cap）的那条叫做redundancy ratio。最小的redundancy ratio是前者最大的车流量的那一条（cap），问minimum redundancy ratio是多少。 思路：一开始想通过每次找增广路径时去找出最小的残余量，然后再找最大的，后来发现是错误的。其实在沿着残余网]]>https://blog.youkuaiyun.com/Wall_F/article/details/9898517https://blog.youkuaiyun.com/Wall_F/article/details/9898517Wall_FSun, 11 Aug 2013 11:53:42 +0800https://blog.youkuaiyun.com/Wall_F/article/details/9878567https://blog.youkuaiyun.com/Wall_F/article/details/9878567Wall_FSat, 10 Aug 2013 15:35:00 +0800https://blog.youkuaiyun.com/Wall_F/article/details/9859965https://blog.youkuaiyun.com/Wall_F/article/details/9859965Wall_FFri, 09 Aug 2013 17:39:21 +0800 0）有流流过，则把该边反向，如果没有，则不改，然后用套圈算法去找欧拉回路。 /*UVA 10735*/ #include #include #include #include #include #in]]>https://blog.youkuaiyun.com/Wall_F/article/details/9664729https://blog.youkuaiyun.com/Wall_F/article/details/9664729Wall_FWed, 31 Jul 2013 10:48:11 +0800https://blog.youkuaiyun.com/Wall_F/article/details/9622241https://blog.youkuaiyun.com/Wall_F/article/details/9622241Wall_FMon, 29 Jul 2013 21:28:45 +0800https://blog.youkuaiyun.com/Wall_F/article/details/9614163https://blog.youkuaiyun.com/Wall_F/article/details/9614163Wall_FMon, 29 Jul 2013 11:46:14 +0800https://blog.youkuaiyun.com/Wall_F/article/details/9614135https://blog.youkuaiyun.com/Wall_F/article/details/9614135Wall_FMon, 29 Jul 2013 11:45:16 +0800