What Is a Bigger Smarter? (Longest DAG path)

The attributes of n elephants, including weight w and IQ s, are given. Now we need to find a continuous sequence, requiring each elephant to have a higher weight than the previous one, the IQ is small, and the maximum value and solution are output. If the solution is not unique, either of them is output. Solution: The longest path of DAG without a fixed point, memory-based search, and record the precursor of the most current solution.

#include <stdio.h>  #include <string.h>  const int N = 10005;    struct State {      int w;      int s;  }tmp[N];  int n, dp[N], vis[N];    int find(int cur) {      if (dp[cur])    return dp[cur];      int a;      vis[cur] = cur;      State& now = tmp[cur];      for (int i = 0; i < n; i++) {      if (tmp[i].w < now.w && tmp[i].s > now.s) {          a = find(i);          if (dp[cur] <= a) {          dp[cur] = a;          vis[cur] = i;          }      }      }      return ++dp[cur];  }    void solve() {      int Max = 0, id = 0, ans[N], cnt = 0;      for (int i = 0; i < n; i++) {      if (!dp[i]) find(i);      if (Max < dp[i]) {          Max = dp[i];          id = i;      }      }        printf("%d\n", Max);      while (id != vis[id]) {      ans[cnt++] = id + 1;      id = vis[id];      }      ans[cnt++] = id + 1;        for (int i = cnt - 1; i >= 0; i--)      printf("%d\n", ans[i]);    }    int main() {      // Init;      n = 0;      memset(tmp, 0, sizeof(tmp));      memset(vis, 0, sizeof(vis));      memset(dp, 0, sizeof(dp));        // Read;      while(scanf("%d%d", &tmp[n].w, &tmp[n].s) == 2)      n++;        solve();        return 0;  }