hdu 1160 FatMouse's Speed
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题目描述:
FatMouse's Speed
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5675 Accepted Submission(s): 2444
Special Judge
Problem Description
FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection of mice and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the speeds are decreasing.
Input
Input contains data for a bunch of mice, one mouse per line, terminated by end of file.
The data for a particular mouse will consist of a pair of integers: the first representing its size in grams and the second representing its speed in centimeters per second. Both integers are between 1 and 10000. The data in each test case will contain information for at most 1000 mice.
Two mice may have the same weight, the same speed, or even the same weight and speed.
The data for a particular mouse will consist of a pair of integers: the first representing its size in grams and the second representing its speed in centimeters per second. Both integers are between 1 and 10000. The data in each test case will contain information for at most 1000 mice.
Two mice may have the same weight, the same speed, or even the same weight and speed.
Output
Your program should output a sequence of lines of data; the first line should contain a number n; the remaining n lines should each contain a single positive integer (each one representing a mouse). If these n integers are m[1], m[2],..., m[n] then it must be the case that
W[m[1]] < W[m[2]] < ... < W[m[n]]
and
S[m[1]] > S[m[2]] > ... > S[m[n]]
In order for the answer to be correct, n should be as large as possible.
All inequalities are strict: weights must be strictly increasing, and speeds must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one.
W[m[1]] < W[m[2]] < ... < W[m[n]]
and
S[m[1]] > S[m[2]] > ... > S[m[n]]
In order for the answer to be correct, n should be as large as possible.
All inequalities are strict: weights must be strictly increasing, and speeds must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one.
Sample Input
6008 13006000 2100500 20001000 40001100 30006000 20008000 14006000 12002000 1900
Sample Output
44597
Source
Zhejiang University Training Contest 2001
思路:
简单dp。。。
dp[i]表示第i只mouse在解里面的话他的最大值;
有状态转移方程:
dp[i]=max(dp[j]+1) (0<=j<i,且j满足条件)
记录每个最优点的父亲,回溯输出解。。。
附上代码:
思路:
简单dp。。。
dp[i]表示第i只mouse在解里面的话他的最大值;
有状态转移方程:
dp[i]=max(dp[j]+1) (0<=j<i,且j满足条件)
记录每个最优点的父亲,回溯输出解。。。
附上代码:
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<set>#include<cstdlib>#define BUG puts("ok");#define CLR(a,b) memset((a),(b),sizeof((a)))#define inf 16843009using namespace std;int const nMax=1010;typedef long long LL;struct Int{ int w,s,tex; bool operator < (const Int & b)const { if(w==b.w)return s>b.s; return w<b.w; }};Int a[nMax];int dp[nMax];int fa[nMax];int n;void output(int u){ if(fa[u]==u){ printf("%d\n",a[u].tex); return ; } output(fa[u]); printf("%d\n",a[u].tex); return ;}int main(){ n=0; while(~scanf("%d%d",&a[n].w,&a[n].s)&&(a[n].tex=n+1))n++; sort(a,a+n); CLR(fa,-1); for(int i=0;i<n;i++){ dp[i]=1;fa[i]=i; for(int j=0;j<i;j++)if(a[j].w<a[i].w&&a[j].s>a[i].s){ if(dp[j]+1>dp[i]){ dp[i]=dp[j]+1; fa[i]=j; } } } int ct=0; for(int i=0;i<n;i++)if(dp[ct]<dp[i])ct=i; printf("%d\n",dp[ct]); output(ct); return 0;}
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