SRM474

250pt

题意：在一个N维的空间里，有一个人开始在原点，现在给出N<=50个指令序列，每个指令序列为某一维+1或者减一，问是否经过某个点至少2次。

思路：操作很小，直接模拟判断即可

code：

 #line 7 "RouteIntersection.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <iostream>
 #include <sstream>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <utility>
 #include <ctime>
 using namespace std;

 #define PB push_back
 #define MP make_pair

 #define REP(i,n) for(i=0;i<(n);++i)
 #define FOR(i,l,h) for(i=(l);i<=(h);++i)
 #define FORD(i,h,l) for(i=(h);i>=(l);--i)

 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef long long LL;
 typedef pair<int,int> PII;

 bool cmp(const pair<int, int>& a, const pair<int,int>& b){
               return abs(a.second) > abs(b.second);
 }

 class RouteIntersection
 {
         public:
         vector< PII > P[];
         bool equal(int a,int b){
               if (P[a].size() != P[b].size()) return false;
               for (int i = ; i < P[a].size(); ++i)
                 if (P[a] != P[b]) return false;
               return true;
         }
         string isValid(int N, vector <int> coords, string moves)
         {
                int m = coords.size();
                P[].clear();
                for (int i = ; i <= m; ++i){
                      int p = -, x = coords[i-], y;
                      if (moves[i-] == '+') y = ;
                      else y = -;
                      P[i] = P[i-];
                      for (int j = ; j < P[i].size(); ++j)
                         if (P[i][j].first == x) p = j;
                      if (p == -) P[i].push_back(make_pair(x, y));
                      else  P[i][p].second += y;
                }
                for (int i = ; i <= m; ++i){
                     sort(P[i].begin(), P[i].end(), cmp);
                     int sz = P[i].size();
                     while (sz > )
                            if (P[i][--sz].second == ) P[i].pop_back();
                            else break;
                     sort(P[i].begin(), P[i].end());
                }
                for (int i = ; i <= m; ++i){
                   //  printf("%d\n", P[i].size());
                    // for (int j = 0; j < P[i].size(); ++j)
                   //       printf("a = %d b = %d   ", P[i][j].first, P[i][j].second);
                 //    puts("");
                     if (!P[i].size()) return "NOT VALID";
                     for (int j = i+; j <= m; ++j)
                        if (equal(i, j)) return  "NOT VALID";
                }
                return "VALID";
         }
 };

500pt

题意：题目给定N<=50的无向连通图，现要你生成一个生成树，并且满足每个点到0的距离正好为原图0到该点的最短路距离。求方案数。

思路：先求一边由0点出发的spfa，并统计每个点的最短路前驱有几个，接着乘法原理即可。

code：

 #line 7 "TreesCount.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <iostream>
 #include <sstream>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <utility>
 #include <ctime>
 using namespace std;
 #define PB push_back
 #define MP make_pair
 #define Inf 0x3fffffff
 #define REP(i,n) for(i=0;i<(n);++i)
 #define FOR(i,l,h) for(i=(l);i<=(h);++i)
 #define FORD(i,h,l) for(i=(h);i>=(l);--i)
 #define M 1000000007
 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef long long LL;
 typedef pair<int,int> PII;

 class TreesCount
 {
         public:
         int d[], dg[];
         bool v[];
         int count(vector <string> S)
         {
               int n = S.size();
               memset(dg,  , sizeof(dg));
               memset(v, , sizeof(v));
               for (int i = ; i < n; ++i) d[i] = Inf;
               queue<int> q;
               q.push();
               d[] = ;
               dg[] = ;
               v[] = true;
               int x, dst;
               while (!q.empty()){
                    x = q.front();
                    for (int y = ; y < n; ++y){
                         dst = S[x][y] - '';
                         if (dst >  && d[x] + dst <= d[y]){
                               if (d[x] + dst < d[y]){
                                       dg[y] = ;
                                       d[y] = d[x] + dst;
                                       if (!v[y]) q.push(y), v[y] = true;
                               }
                               else ++dg[y];
                         }
                    }
                    v[x] = false;
                    q.pop();
               }
               long long ans = ;
               for (int i = ; i < n; ++i)
                  ans =  (ans * dg[i]) % M;
               return ans;
         }
 };