Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

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] 表示到第 i 站的剩餘 j 油量。
最簡單的方程 dp[i][j] = min(dp[i-1][j-dis((i-1)->(i))], min(dp[i][j-k]+k*p[i]))

想一想，真的需要枚舉 k 嗎？一次加 k 的油量？

其實只要跟 dp[i][j-1] + p[i] 就可以了，跟前一個狀態比加一升就可以了。

#include <stdio.h>
#include <algorithm>
#define oo 100000000
using namespace std;

int main() {
    int t, D;
    int i, j, dp[105][205];
    char s[105];
    scanf("%d", &t);
    while(getchar() != '\n');
    gets(s);
    while(t--) {
        scanf("%d", &D);
        while(getchar() != '\n');
        int d[105], p[105], n = 1;
        d[0] = 0, p[0] = 0;
        while(gets(s) && s[0]) {
            sscanf(s, "%d %d", &d[n], &p[n]);
            n++;
        }
        d[n] = D, p[n] = oo;
        for(i = 0; i <= n; i++)
            for(j = 0; j <= 200; j++)
                dp[i][j] = oo;
        dp[0][100] = 0;
        for(i = 1; i <= n; i++) {
            int dis = d[i]-d[i-1];
            for(j = dis; j <= 200; j++)
                dp[i][j-dis] = min(dp[i][j-dis], dp[i-1][j]);
            for(j = 1; j <= 200; j++)
                dp[i][j] = min(dp[i][j], dp[i][j-1]+p[i]);
        }
        int ans = oo;
        for(i = 100; i <= 200; i++)
            ans = min(ans, dp[n][i]);
        if(ans == oo)
            puts("Impossible");
        else
            printf("%d\n", ans);
        if(t)
            puts("");
    }
    return 0;
}