CodeForces 754C Vladik and chat (DP+暴力)

题意：给定n个人的m个对话，问能不能找一个方式使得满足，上下楼层人名不同，并且自己不提及自己。

析：首先预处理每一层能有多少个user可选，dp[i][j] 表示第 i 层是不是可以选第 j 个user。最后再输出即可。

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 <stack>
#include <sstream>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1LL << 60;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
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 bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
map<string, int> mp;
int cnt;
vector<string> v;
bool vis[maxn][maxn];
int dp[maxn][maxn], a[maxn];
string name[maxn];

inline int getId(const string &s){
    if(!mp.count(s))  name[mp[s] = cnt++] = s;
    return mp[s];
}

string getF(const string &t){
    string s = "";
    for(int j = 0; t[j] != ':'; ++j)  s.push_back(t[j]);
    return s;
}

void calc1(bool *is, string s){
    for(int i = 0; i < s.size(); ++i)
        if(ispunct(s[i]))  s[i] = ' ';
   // cout << s << endl;
    stringstream ss(s);
    while(ss >> s) if(mp.count(s))  is[mp[s]] = true;
}

int main(){
    ios::sync_with_stdio(false);
    int T;  cin >> T;
    while(T--){
        cin >> n;
        mp.clear();
        cnt = 1;
        string s;
        memset(vis, 0, sizeof vis);
        for(int i = 0; i < n; ++i){
            cin >> s;
            getId(s);
        }
        cin >> m;
        cin.get();
        v.clear();
        v.push_back(" ");
        for(int i = 1; i <= m; ++i){
            getline(cin, s);
            v.push_back(s);
            calc1(vis[i], s);
            a[i] = v[i][0] == '?' ? 0 : mp[getF(s)];
        }

        memset(dp, 0, sizeof dp);
        int tmp = 1;
        dp[0][0] = 1;
        for(int i = 1; i <= m; ++i){
            int tt = 0;
            if(a[i]){
                dp[i][a[i]] = tmp > dp[i-1][a[i]];
                tt = dp[i][a[i]];
            }
            else for(int j = 1; j <= n; ++j){
                dp[i][j] = (!vis[i][j] && tmp > dp[i-1][j]);
                tt += dp[i][j];
            }
            tmp = tt;
        }
        tmp = 0;
        for(int i = 1; i <= n; ++i)  if(dp[m][i]){ tmp = i;  break;  }
        if(!tmp)  cout << "Impossible" << endl;
        else{
            for(int i = m; i > 0; --i)
                for(int j = 0; j <= n; ++j)
                    if(j != tmp && dp[i-1][j]){
                        if(!a[i])  a[i] = -tmp;
                        tmp = j;
                        break;
                    }
            for(int i = 1; i <= m; ++i)
                if(a[i] > 0)  cout << v[i] << endl;
                else cout << name[-a[i]] + v[i].substr(1) << endl;
        }
    }
    return 0;
}