HDU 1160 FatMouse's Speed

FatMouse's Speed
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14892 Accepted Submission(s): 6565
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.

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.

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

Source [Zhejiang University Training Contest 2001](http://acm.hdu.edu.cn/search.php?field=problem&key=Zhejiang+University+Training+Contest+2001&source=1&searchmode=source)
解析：最长上升子序列的加强版。对于本题，我们可以做一个预处理，按每只老鼠的重量从小到大排序。接着速度就变成了一个最长上升子序列的模型。不过本题速度应该按照下降来计算。注意两只老鼠的重量和速度都可以相同，判断的时候要同时判断。并在状态转移的时候用pre记录子序列路径各个位置的前驱，最后倒序输出即可。
``` #include #include #include using namespace std;

struct M{
int w, s, id;
bool operator < (const M &b)const
{
return w < b.w;
}
};
M m[1005];

struct DP{
int val, pre;
bool operator < (const DP &b)const
{
return val < b.val;
}
};
DP dp[1005];

void solve(int n)
{
sort(m, m+n);
for(int i = 0; i < n; ++i){
dp[i].val = 1; dp[i].pre = -1;
for(int j = 0; j < i; ++j){
if(m[j].w < m[i].w && m[j].s > m[i].s){
if(dp[j].val+1 > dp[i].val){
dp[i].val = dp[j].val+1;
dp[i].pre = j;
}
}
}
}
auto it = max_element(dp, dp+n);
printf("%d\n", it->val);
int d = it-dp;
stack st;
while(d != -1){
st.push(m[d].id);
d = dp[d].pre;
}
while(!st.empty()){
printf("%d\n", st.top());
st.pop();
}
}

int main()
{
int cnt = 0;
while(~scanf("%d%d", &m[cnt].w, &m[cnt].s)){
m[cnt].id = cnt+1;
++cnt;
}
solve(cnt);
return 0;
}