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BZOJ3315: [Usaco2013 Nov]Pogo-Cow
3315: [Usaco2013 Nov]Pogo-Cow
Time Limit: 3 Sec Memory Limit: 128 MBSubmit: 143 Solved: 79
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Description
In an ill-conceived attempt to enhance the mobility of his prize cow Bessie, Farmer John has attached a pogo stick to each of Bessie‘s legs. Bessie can now hop around quickly throughout the farm, but she has not yet learned how to slow down. To help train Bessie to hop with greater control, Farmer John sets up a practice course for her along a straight one-dimensional path across his farm. At various distinct positions on the path, he places N targets on which Bessie should try to land (1 <= N <= 1000). Target i is located at position x(i), and is worth p(i) points if Bessie lands on it. Bessie starts at the location of any target of her choosing and is allowed to move in only one direction, hopping from target to target. Each hop must cover at least as much distance as the previous hop, and must land on a target. Bessie receives credit for every target she touches (including the initial target on which she starts). Please compute the maximum number of points she can obtain.
一个坐标轴有N个点，每跳到一个点会获得该点的分数，并只能朝同一个方向跳，但是每一次的跳跃的距离必须不小于前一次的跳跃距离，起始点任选，求能获得的最大分数。
Input
* Line 1: The integer N.
* Lines 2..1+N: Line i+1 contains x(i) and p(i), each an integer in the range 0..1,000,000.
Output
* Line 1: The maximum number of points Bessie can receive.
Sample Input
5 6
1 1
10 5
7 6
4 8
8 10
INPUT DETAILS: There are 6 targets. The first is at position x=5 and is worth 6 points, and so on.
Sample Output
OUTPUT DETAILS: Bessie hops from position x=4 (8 points) to position x=5 (6 points) to position x=7 (6 points) to position x=10 (5 points).
从坐标为4的点，跳到坐标为5的，再到坐标为7和，再到坐标为10的。
HINT
Source
Silver
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set>10 #include<queue>11 #include<string>12 #define inf 100000000013 #define maxn 150014 #define maxm 500+10015 #define eps 1e-1016 #define ll long long17 #define pa pair<int,int>18 #define for0(i,n) for(int i=0;i<=(n);i++)19 #define for1(i,n) for(int i=1;i<=(n);i++)20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)22 #define mod 100000000723 using namespace std;24 inline int read()25 {26 int x=0,f=1;char ch=getchar();27 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}28 while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}29 return x*f;30 }31 int n,ans=0,f[maxn][maxn];32 struct rec{int x,y;}a[maxn];33 inline bool cmp(rec a,rec b)34 {35 return a.x<b.x;36 }37 int main()38 {39 freopen("input.txt","r",stdin);40 freopen("output.txt","w",stdout);41 n=read();42 for1(i,n)a[i].x=read(),a[i].y=read();43 sort(a+1,a+n+1,cmp);44 a[0].x=-inf;45 for1(i,n)46 {47 f[i][i]=a[i].y;48 int k=i+1,tmp=f[i][i];49 for3(j,i-1,0)50 {51 while(k<=n&&a[k].x-a[i].x<a[i].x-a[j].x)52 {53 f[k][i]=max(f[k][i],tmp+a[k].y);54 //cout<<k<<‘ ‘<<i<<‘ ‘<<f[k][i]<<endl;55 ans=max(ans,f[k][i]);56 k++;57 }58 tmp=max(tmp,f[i][j]);59 if(k>n)break;60 }61 }62 //for1(i,n)for1(j,i-1)cout<<i<<‘ ‘<<j<<‘ ‘<<f[i][j]<<endl;63 memset(f,0,sizeof(f));64 a[n+1].x=inf;65 for3(i,n,1)66 {67 f[i][i]=a[i].y;68 int k=i-1,tmp=f[i][i];69 for2(j,i+1,n+1)70 {71 while(k&&a[i].x-a[k].x<a[j].x-a[i].x)72 {73 f[k][i]=max(f[k][i],tmp+a[k].y);74 //cout<<k<<‘ ‘<<i<<‘ ‘<<tmp<<‘ ‘<<f[k][i]<<endl;75 ans=max(ans,f[k][i]);76 k--;77 }78 tmp=max(tmp,f[i][j]);79 if(!k)break;80 }81 }82 printf("%d\n",ans); 83 return 0;84 }
BZOJ3315: [Usaco2013 Nov]Pogo-Cow