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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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原文地址：http://www.cnblogs.com/dwtfukgv/p/5936811.html
