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：
1 #line 7 "RouteIntersection.cpp" 2 #include <cstdlib> 3 #include <cctype> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <vector> 9 #include <string> 10 #include <iostream> 11 #include <sstream> 12 #include <map> 13 #include <set> 14 #include <queue> 15 #include <stack> 16 #include <fstream> 17 #include <numeric> 18 #include <iomanip> 19 #include <bitset> 20 #include <list> 21 #include <stdexcept> 22 #include <functional> 23 #include <utility> 24 #include <ctime> 25 using namespace std; 26 27 #define PB push_back 28 #define MP make_pair 29 30 #define REP(i,n) for(i=0;i<(n);++i) 31 #define FOR(i,l,h) for(i=(l);i<=(h);++i) 32 #define FORD(i,h,l) for(i=(h);i>=(l);--i) 33 34 typedef vector<int> VI; 35 typedef vector<string> VS; 36 typedef vector<double> VD; 37 typedef long long LL; 38 typedef pair<int,int> PII; 39 40 bool cmp(const pair<int, int>& a, const pair<int,int>& b){ 41 return abs(a.second) > abs(b.second); 42 } 43 44 class RouteIntersection 45 { 46 public: 47 vector< PII > P[65]; 48 bool equal(int a,int b){ 49 if (P[a].size() != P[b].size()) return false; 50 for (int i = 0; i < P[a].size(); ++i) 51 if (P[a] != P[b]) return false; 52 return true; 53 } 54 string isValid(int N, vector <int> coords, string moves) 55 { 56 int m = coords.size(); 57 P[0].clear(); 58 for (int i = 1; i <= m; ++i){ 59 int p = -1, x = coords[i-1], y; 60 if (moves[i-1] == '+') y = 1; 61 else y = -1; 62 P[i] = P[i-1]; 63 for (int j = 0; j < P[i].size(); ++j) 64 if (P[i][j].first == x) p = j; 65 if (p == -1) P[i].push_back(make_pair(x, y)); 66 else P[i][p].second += y; 67 } 68 for (int i = 1; i <= m; ++i){ 69 sort(P[i].begin(), P[i].end(), cmp); 70 int sz = P[i].size(); 71 while (sz > 0) 72 if (P[i][--sz].second == 0) P[i].pop_back(); 73 else break; 74 sort(P[i].begin(), P[i].end()); 75 } 76 for (int i = 1; i <= m; ++i){ 77 // printf("%d\n", P[i].size()); 78 // for (int j = 0; j < P[i].size(); ++j) 79 // printf("a = %d b = %d ", P[i][j].first, P[i][j].second); 80 // puts(""); 81 if (!P[i].size()) return "NOT VALID"; 82 for (int j = i+1; j <= m; ++j) 83 if (equal(i, j)) return "NOT VALID"; 84 } 85 return "VALID"; 86 } 87 };
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：
1 #line 7 "TreesCount.cpp" 2 #include <cstdlib> 3 #include <cctype> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <vector> 9 #include <string> 10 #include <iostream> 11 #include <sstream> 12 #include <map> 13 #include <set> 14 #include <queue> 15 #include <stack> 16 #include <fstream> 17 #include <numeric> 18 #include <iomanip> 19 #include <bitset> 20 #include <list> 21 #include <stdexcept> 22 #include <functional> 23 #include <utility> 24 #include <ctime> 25 using namespace std; 26 #define PB push_back 27 #define MP make_pair 28 #define Inf 0x3fffffff 29 #define REP(i,n) for(i=0;i<(n);++i) 30 #define FOR(i,l,h) for(i=(l);i<=(h);++i) 31 #define FORD(i,h,l) for(i=(h);i>=(l);--i) 32 #define M 1000000007 33 typedef vector<int> VI; 34 typedef vector<string> VS; 35 typedef vector<double> VD; 36 typedef long long LL; 37 typedef pair<int,int> PII; 38 39 40 class TreesCount 41 { 42 public: 43 int d[120], dg[120]; 44 bool v[120]; 45 int count(vector <string> S) 46 { 47 int n = S.size(); 48 memset(dg, 0 , sizeof(dg)); 49 memset(v, 0, sizeof(v)); 50 for (int i = 0; i < n; ++i) d[i] = Inf; 51 queue<int> q; 52 q.push(0); 53 d[0] = 0; 54 dg[0] = 1; 55 v[0] = true; 56 int x, dst; 57 while (!q.empty()){ 58 x = q.front(); 59 for (int y = 0; y < n; ++y){ 60 dst = S[x][y] - '0'; 61 if (dst > 0 && d[x] + dst <= d[y]){ 62 if (d[x] + dst < d[y]){ 63 dg[y] = 1; 64 d[y] = d[x] + dst; 65 if (!v[y]) q.push(y), v[y] = true; 66 } 67 else ++dg[y]; 68 } 69 } 70 v[x] = false; 71 q.pop(); 72 } 73 long long ans = 1; 74 for (int i = 0; i < n; ++i) 75 ans = (ans * dg[i]) % M; 76 return ans; 77 } 78 };
