题目链接：?pid=1160
FatMouse’s SpeedTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9965Accepted Submission(s): 4427Special 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.
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 thatW[m[1]] < W[m[2]] < … < W[m[n]]andS[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
最长上升子序列及记录路径
#include<stdio.h>#include<iostream>#include<math.h>#include<stdlib.h>#include<ctype.h>#include<algorithm>#include<vector>#include<string>#include<queue>#include<stack>#include<set>#include<map>using namespace std;struct node{int x, y, num;}p[10010];bool cmp(node a, node b){return a.x < b.x;}int pre[10010], dp[10010];vector<int> q;int main(){int x, y;int len = 1;while (scanf("%d%d",&x,&y)!=EOF)//for (int i = 1; i < 10; i++){//cin >> x >> y;p[len].x = x;p[len].y = y;p[len].num = len;len++;}sort(p+1,p+len,cmp);memset(pre,0,sizeof(pre));for (int i = 1; i < len; i++){dp[i] = 1;for (int j = 1; j < i; j++){if (p[i].x > p[j].x && p[i].y < p[j].y ){if (dp[i] < dp[j] + 1){dp[i] = dp[j] + 1;pre[p[i].num] = p[j].num;}}}}int ans = -1, num = 1;for (int i = 1; i < len; i++){if (ans < dp[i]){ans = dp[i];num = p[i].num;}}printf("%d\n",ans);q.clear();while (num != 0){q.push_back(num);num = pre[num];}reverse(q.begin(),q.end());for (int i = 0; i < q.size(); i++)printf("%d\n",q[i]);return 0;}
，继续期待我的下一个旅行，拿起背包，
