HDU1160 FatMouse's Speed —— DP
题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=1160
FatMouse's Speed
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17386 Accepted Submission(s): 7694
Special Judge
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.
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.
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900
4
5
9
7
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e3+; struct node
{
int wei, spd, id;
bool operator<(const node &x) {
return wei<x.wei;
}
}mice[MAXN]; int dp[MAXN], pre[MAXN]; void Print(int k) //输出路径
{
if(!k) return;
Print(pre[k]);
printf("%d\n", mice[k].id); //由于经过了排序,所以输出的是原始编号。
} int main()
{
int n = ;
int wei, spd;
while(scanf("%d%d", &wei, &spd)!=EOF)
{
mice[++n].wei = wei;
mice[n].spd = spd;
mice[n].id = n;
} sort(mice+, mice++n);
mice[].wei = -INF, mice[].spd = INF; //!!注意边界条件
memset(dp, , sizeof(dp));
memset(pre, , sizeof(pre)); int k = -;
for(int i = ; i<=n; i++)
for(int j = ; j<i; j++) //下标从0开始, 表明i作为第一个
{
if(mice[i].wei>mice[j].wei && mice[i].spd<mice[j].spd && dp[i]<dp[j]+)
{
dp[i] = dp[j]+;
pre[i] = j; //同时更新pre, 用于输出路径
}
if(k==- || dp[i]>dp[k])
k = i;
} printf("%d\n", dp[k]);
Print(k);
}
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