UVaLive 6680 Join the Conversation (DP)
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题意:给出n条发言,让你求最大的交流长度并输出标记顺序。
析:这个题要知道的是,前面的人是不能at后面的人,只能由后面的人at前面的,那就简单了,我们只要更新每一层的最大值就好,并不会影响到其他层。
最后再从这 n 层中取出最大值,在更新时,也可以记录着最大值。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <sstream> #include <stack> //#include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 5e4 + 5; const int mod = 1e9 + 7; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } map<string, P> mp; int dp[maxn], pre[maxn]; void print(int pos){ if(!pre[pos]){ printf("%d", pos); return; } print(pre[pos]); printf(" %d", pos); } int main(){ while(scanf("%d", &n) == 1){ string line, s; getchar(); mp.clear(); for(int i = 1; i <= n; ++i){ getline(cin, line); stringstream ss(line); ss >> s; string str = s.substr(0, s.size()-1); dp[i] = 1, pre[i] = 0; while(ss >> s){ if(!mp.count(s) || s == str) continue; if(mp[s].first + 1 > dp[i]){ dp[i] = mp[s].first + 1; pre[i] = mp[s].second; } } if(mp[str].first < dp[i]) mp[str] = P(dp[i], i); } int ans = 0, pos; for(int i = 1; i <= n; ++i) if(dp[i] > ans) ans = dp[i], pos = i; printf("%d\n", ans); print(pos); printf("\n"); } return 0; }
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