Input In the first line there is an integer t ( 1≤t≤50), indicating the number of test cases.For each test case:The first line contains four integers, n, A, B, L.Next n lines, each line contains two integers: Li,Ri, which represents the interval [Li,Ri] is swamp.1≤n≤100,1≤L≤105,1≤A≤10,1≤B≤10，1≤Li<Ri≤L.Make sure intervals are not overlapped which means Ri<Li+1 for each i (1≤i<n).Others are all flats except the swamps.
Output For each text case: Please output “Case #k: answer”(without quotes) one line, where k means the case number counting from 1, and the answer is his minimum strengths in the beginning.
Sample Input 1 2 2 2 5 1 2 3 4
Sample Output Case #1: 0
Source 2015 ACM/ICPC Asia Regional Shanghai Online 题意：要我们找需要消耗体力最多的那个地方（即你走一段沼泽，可能会体力为0，这就是你开头需要准备的体力。或是你上一段体力补充完了，这一段又被沼泽消耗光了体力，这也是要补充的体力）。读不懂题真要命 #include <cstdio> #include <iostream> #include <cmath> #include <string> #include <cstring> #include <algorithm> #include <queue> #include <vector> #include <map> using namespace std; #define ll long long const int maxn = 5100; int t, n, A, B, L, l, r; int main() { scanf("%d", &t); int miao = t; while(t--) { scanf("%d%d%d%d", &n, &A, &B, &L); int legacy = 0, position = 0, need = 0;//legacy是剩余耐力，need是需求，position是当前位置 for(int i = 0; i<n; i++) { scanf("%d%d", &l, &r); legacy += (l-position)*B-(r-l)*A;//（上一段平底回复的）-（这一段澡泽消耗的） position = r;//把上一段澡泽的最后一个位置标记下来 if(legacy<0)//耐力不够时需要额外添加 { need -= legacy;//这时候legacy是负的，所以要减 legacy = 0; } } printf("Case #%d: %d\n", miao-t, need); } return 0; }
转载于:https://www.cnblogs.com/RootVount/p/11235961.html
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