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. 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 m11, m22,…, mnn then it must be the case that
Wm[1m[1] < Wm[2m[2] < … < Wm[nm[n]
and
Sm[1m[1] > Sm[2m[2] > … > Sm[nm[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 1300 6000 2100 500 2000 1000 4000 1100 3000 6000 2000 8000 1400 6000 1200 2000 1900 Sample Output 4 4 5 9 7
題目大意是找到一個最多的老鼠序列，使得序列中的老鼠的體重滿足遞增，相應老鼠的速度滿足遞減。 這種要和兩個因素排列有關的一般都是先排一個因素，再dp另一個因素 這就很簡單了 for i 1~n for j 1~i-1 a[i]>a[j] dp[i]=max(dp[i],dp[j]+1) ; 再用一個pre的數組記錄路徑。記錄最大的點，然后往回推。 int t=maxi; i=0; while(t!=0) { res[i++]=t; t=pre[t]; }
#include<stdio.h>#include<algorithm>using namespace std;#define MAXN 1000struct Node{ int w,s;//重量和速度 int index;//最初的序號，避免排序后亂掉順序，后面需要輸出的 }mouse[MAXN+10];bool cmp(Node a,Node b)//先按照w從小到大排序，再按照y從大到小排序 { if(a.w<b.w) return 1; else if(a.w==b.w&&a.s>b.s)return 1; else return 0;} int dp[MAXN+10];//dp[i]表示以第i個數據結尾的符合要求的子列長度int pre[MAXN+10];//記錄i對應的上一個數據int res[MAXN+10];//存放最終結果下標 int main(){ //freopen("test.in","r",stdin); //freopen("test.out","w",stdout); int i=1,j; while(scanf("%d%d",&mouse[i].w,&mouse[i].s)!=EOF) { dp[i]=1; pre[i]=0; mouse[i].index=i; i++; } int n=i-1; sort(mouse+1,mouse+1+n,cmp); int maxlen=0;//最長序列長度 int maxi;//最長序列的最后一個數下標 dp[1]=1; for(i=1;i<=n;i++) { for(j=1;j<i;j++) if(mouse[i].w>mouse[j].w&&mouse[i].s<mouse[j].s&&dp[j]+1>dp[i]) { dp[i]=dp[j]+1; pre[i]=j; if(dp[i]>maxlen) { maxi=i; maxlen=dp[i]; } } } int t=maxi; i=0; while(t!=0) { res[i++]=t; t=pre[t]; } printf("%d/n",i); while(i>0) { i--; printf("%d/n",mouse[res[i]].index); } return 0; }新聞熱點
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