H - Adventures in Moving - Part IV
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
Submit Status Practice UVA 10201
Description
Problem A: Adventures in Moving - Part IV
To help you move from Waterloo to the big city, you are considering renting a moving truck. Gas prices being so high these days, you want to know how much the gas for such a beast will set you back.
The truck consumes a full litre of gas for each kilometre it travels. It has a 200 litre gas tank. When you rent the truck in Waterloo, the tank is half full. When you return it in the big city, the tank must be at least half full, or you‘ll get gouged even more for gas by the rental company. You would like to spend as little as possible on gas, but you don‘t want to run out along the way.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
Input is all integers. The first integer is the distance in kilometres from Waterloo to the big city, at most 10000. Next comes a set of up to 100 gas station specifications, describing all the gas stations along your route, in non-decreasing order by distance. Each specification consists of the distance in kilometres of the gas station from Waterloo, and the price of a litre of gas at the gas station, in tenths of a cent, at most 2000.
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
Output is the minimum amount of money that you can spend on gas to get you from Waterloo to the big city. If it is not possible to get from Waterloo to the big city subject to the constraints above, print "Impossible".
Sample Input
1 500 100 999 150 888 200 777 300 999 400 1009 450 1019 500 1399
Output for Sample Input
450550
dp[i][j]表示到达第i个加油站剩余油量为j时的最小花费
特殊地，dp[n][j]表示到达终点剩余油量为j时的最小花费
状态转移：(设w[i]为每个加油站的位置，p[i]为油单价)
行驶：dp[i][j-(w[i]-w[i-1])] = dp[i-1][j](0<i<=n)
加油：dp[i][j] = dp[i][j-1]+p[i](0<i<n)因为n是特意增加的终点，不是加油站