HDU 杭电5477 A Sweet Journey
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http://acm.hdu.edu.cn/showproblem.php?pid=5477
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 meansRi<Li+1 for each i (1≤i<n ).
Others are all flats except the swamps.
For each test case:
The first line contains four integers, n, A, B, L.
Next n lines, each line contains two integers:
Make sure intervals are not overlapped which means
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.
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
//通俗易懂,纯暴力
#include <stdio.h>#include <string.h>#define min(a,b) (a<b?a:b)#define INF 0x3fffffffint dp[100110];int main(){int T, N;int a, b;int A, B;int L;int cas=0;scanf("%d",&T);while(T--){memset(dp,0,sizeof(dp));int res=0;scanf("%d%d%d%d",&N,&A,&B,&L);for(int i=0;i<N;++i){scanf("%d%d",&a,&b);for(int j=a;j<b;++j)dp[j]=1;}int ans=INF;for(int i=0;i<L;++i){if(dp[i]) {res-=A;ans=min(res,ans);}else res+=B;}printf("Case #%d: ",++cas);if(ans>=0) printf("0\n");else printf("%d\n",-ans);}return 0;}
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