https://blog.youkuaiyun.com/pai_pkzh-cnhttps://blog.youkuaiyun.com/https://blog.youkuaiyun.com/pai_pk/article/details/52014165https://blog.youkuaiyun.com/pai_pk/article/details/52014165pai_pkSun, 24 Jul 2016 15:27:04 +0800（或(用"gt”表示)或 思路：简单的差分约束，就是题目描述有点不太好懂。需要注意，虽然序列长度为n，但是建图的时候用来第0项，相当于序列长度变为了n+1，所以bellman算法要循环n次！ #include #include #include #include #include #]]>https://blog.youkuaiyun.com/pai_pk/article/details/52013228https://blog.youkuaiyun.com/pai_pk/article/details/52013228pai_pkSun, 24 Jul 2016 11:43:45 +0800= ci;  2. S[i] - S[i-1] = 0; 以mx表示所有区间右端点的最大值，mn表示所有区间左端点的最小值，最后所求结果即为dist[]]>https://blog.youkuaiyun.com/pai_pk/article/details/51853696https://blog.youkuaiyun.com/pai_pk/article/details/51853696pai_pkThu, 07 Jul 2016 19:23:45 +0800https://blog.youkuaiyun.com/pai_pk/article/details/51851940https://blog.youkuaiyun.com/pai_pk/article/details/51851940pai_pkThu, 07 Jul 2016 15:57:42 +0800https://blog.youkuaiyun.com/pai_pk/article/details/51838180https://blog.youkuaiyun.com/pai_pk/article/details/51838180pai_pkWed, 06 Jul 2016 13:38:16 +0800