hdoj 5477 A Sweet Journey 【二分】

A Sweet Journey

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 302    Accepted Submission(s): 163

Problem Description

Master Di plans to take his girlfriend for a travel by bike. Their journey, which can be seen as a line segment of length L, is a road of swamps and flats. In the swamp, it takes A point strengths per meter for Master Di to ride; In the flats, Master Di will regain B point strengths per meter when riding. Master Di wonders：In the beginning, he needs to prepare how much minimum strengths. (Except riding all the time,Master Di has no other choice) 

 

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

12 2 2 51 23 4

 

Sample Output

Case #1: 0

 

题意：一个人要骑车过一个[0, L]的区间，区间里面有n个沼泽其余全是平坦的道路。第i个沼泽的区间为[l[i], r[i]]，所有沼泽都不会重叠且r[i] < l[i+1]。已知骑车过沼泽每单位长度需要花费a点力气，走平路每单位长度可以收获b点力气。问你这个人至少需要准备的力气。。。

本身模拟比较弱，直接上了二分。

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>#define MAXN 100000+1using namespace std;int s[MAXN], e[MAXN];int n, L, a, b;bool judge(int d){    int pre = 0;    for(int i = 1; i <= n; i++)    {        d += (s[i] - pre) * b;        d -= (e[i] - s[i]) * a;        if(d < 0)            return false;        pre = e[i];    }    return true;}int main(){    int t, k = 1;    scanf("%d", &t);    while(t--)    {        scanf("%d%d%d%d", &n, &a, &b, &L);        for(int i = 1; i <= n; i++)            scanf("%d%d", &s[i], &e[i]);        int l = 0, r = L * 10+1, mid;        int ans;        while(r >= l)        {            int mid = (l + r) >> 1;            if(judge(mid))                ans = mid, r = mid - 1;            else                l = mid + 1;        }        printf("Case #%d: %d\n", k++, ans);    }    return 0;}