ZOJ 3689 Digging（DP)
Description
When it comes to the Maya Civilization, we can quickly remind of a term called the end of the world. It's not difficult to understand why we choose to believe the prophecy (or we just assume it is true to entertain ourselves) if you know the other prophecies appeared in the Maya Calendar. For instance, it has accurately predicted a solar eclipse on July 22, 2009.
The ancient civilization, such as Old Babylonianhas, Ancient Egypt and etc, some features in common. One of them is the tomb because of the influence of the religion. At that time, the symbol of the tomb is the pyramid. Many of these structures featured a top platform upon which a smaller dedicatory building was constructed, associated with a particular Maya deity. Maya pyramid-like structures were also erected to serve as a place of interment for powerful rulers.
Now there are N coffin chambers in the pyramid waiting for building and the ruler has recruited some workers to work for T days. It takes ti days to complete the ith coffin chamber. The size of theith coffin chamber is si. They use a very special method to calculate the reward for workers. If starting to build the ith coffin chamber when there are t days left, they can get t*si units of gold. If they have finished a coffin chamber, then they can choose another coffin chamber to build (if they decide to build the ith coffin chamber at the time t, then they can decide next coffin chamber at the time t-ti).
At the beginning, there are T days left. If they start the last work at the time t and the finishing time t-ti < 0, they will not get the last pay.
Input
There are few test cases.
The first line contains N, T (1 ≤ N ≤ 3000,1 ≤ T ≤ 10000), indicating there are N coffin chambers to be built, and there are T days for workers working. Next N lines contains ti, si (1 ≤ ti, si ≤ 500).
All numbers are integers and the answer will not exceed 2^31-1.
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转：http://hi.baidu.com/oldjunyi/item/8391d3e63ae1493986d9decd?qq-pf-to=pcqq.c2c
题意：古代某统治者要修建一些棺材，其中第 i 个棺材大小为 s[i]，修建需要花费 t[i] 天，如果在剩余 x 天的时候开始修建并且能够及时完成，则能获得 x * s[i] 的报酬，总共有 T 天可用，问最大能获得的报酬为多少
先不考虑总天数 T 的限制，假设他们全都能修建完成。对于某个修建顺序 a[1], a[2], .., a[N]，考虑其中任意相邻的两个任务 l = a[i] 和 r = a[i + 1]，它们能获得的报酬为：
x * s[l] + (x - t[l]) * s[r] = x * (s[l] + s[r]) - t[l] * s[r]
如果交换它们的顺序，则明显不影响其他任务（因为它们的总耗时不变），而交换后的报酬为：
x * s[r] + (x - t[r]) * s[l] = x * (s[l] + s[r]) - t[r] * s[l]
可以发现，这两个式子变换后，前面的部分都一样，后面的部分一个是 -t[l] * s[r]，一个是 -t[r] * s[l]
既然交换相邻的任务不会影响其他任务，但会改变的总报酬，那么我们就可以通过对 {1, 2, 3, .., N} 这个序列进行一定的交换，得到一个报酬最大的修建顺序，换句话说，对这些任务进行一个排序即可得到一个最优的工作顺序：
bool cmp(int l, int r){
return -t[l] * s[r] > -t[r] * s[l];
}
现在，有了总天数 T 的限制后，必须在这个基础上进行一个 DP，做法就是从排好序的工作中选择一部分工作去执行（上面的 cmp 函数中的表达式能转换成 t[l] / s[l] < t[r] / s[r]，可以发现它有传递性，因此它的任意子序列也是最优的顺序），
于是剩下的 DP 部分是一个类似于背包的写法，照着题目里给的计算方式去算就行了
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至于排序后的DP和没排序的DP有什么不一样，那是因为根据背包DP的性质，回想一下工作原理可以发现，序号小的物品一定会放在序号大的物品前面，那么就能够符合上述要求了。
