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

           
 A dam has n water gates to let out water when necessary. Each water gate has its own capacity, water path and affected areas in the downstream. The affected areas may have a risk of flood when the water gate is open. The cost of potential damage caused by a water gate is measured in number calculated from the number of people and areas estimated to get affected.

            Suppose a water gate G_i has the volumetric flow rate of F_i m³/hour and the damage cost of C_i. In a certain situation, the dam has the volume V m³ of water to flush out within T hours. Your task is to manage the opening of the water gates in order to get rid of at least the specified volume of water within a limited time in condition that the damage cost is minimized. 

For example, a dam has 4 water gates and their properties are shown in the following table.

Case 1:  You have to flush out the water 5 million m³ within 7 hours. The minimum cost will be 120,000 by letting the water gate G1 open for 7 hours.

Case 2:  You have to flush out the water 5 million m³ within 30 hours. The minimum cost will be 110,000 by letting the water gates G₂ and G₃ open, for example, G₂ is open for 29 hours and G₃ is open for 28 hours.

Note that each water gate is independent and it can be open only in a unit of whole hour (no fraction of hour).

Input

The first line includes an integer n indicating number of water gates (1 ≤ n ≤ 20). Then the next n lines contain, in each line, two integers: F_i and C_i corresponding to the flow rate (m³/hour) and the damage cost of the water gate G_i  respectively. The next line contains the number m which is the number of test cases (1 ≤ m ≤ 50). The following m lines contain, in each line, two integers: V and T corresponding to the volume (m³) of water to let out within T hours.
1 ≤ F_i ,V, C_i ≤ 10⁹ , 1 ≤ T ≤ 1000)

Output

For each test case, print out the minimum cost in the exact format shown in the sample output below. If it is not possible to let out the water of volume V in T hours from the dam, print out “IMPOSSIBLE” (without quotation marks).

DP的話數值太大，那就暴搜吧

#include <stdio.h>
int F[30], C[30];
int ans, V, T, n;
void dfs(int idx, int cost, int vv) {
    if(cost >= ans)
        return;
    if(vv >= V) {
        ans = cost;
        return;
    }
    if(idx == n)
        return;
    dfs(idx+1, cost+C[idx], vv+T*F[idx]);
    dfs(idx+1, cost, vv);
}
int main() {
    int t, i;
    scanf("%d", &n);
    for(i = 0; i < n; i++) {
        scanf("%d %d", &F[i], &C[i]);
    }
    scanf("%d", &t);
    int cases = 0;
    while(t--) {
        scanf("%d %d", &V, &T);
        ans = 2147483647;
        dfs(0,0,0);
        if(ans != 2147483647)
            printf("Case %d: %d", ++cases, ans);
        else
            printf("Case %d: IMPOSSIBLE", ++cases);
        puts("");
    }
    return 0;
}