标签:clu gcd can debug freopen ring queue comm define
题意:让人构造一个图,满足每个结点边的数目不超过 k,然后给出每个结点到某个结点的最短距离。
析:很容易看出来如果可能的话,树是一定满足条件的,只要从头开始构造这棵树就好,中途超了int。。。找了好久。
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> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #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 = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 100000 + 10; const int mod = 100000000; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -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; } vector<int> G[maxn]; int main(){ scanf("%d %d", &n, &m); int rt = -1, cnt = 0; int mmax = 0; for(int i = 1; i <= n; ++i){ int x; scanf("%d", &x); G[x].push_back(i); if(x == 0) rt = i; mmax = max(mmax, x); } if(G[1].size() > G[0].size() * m || G[0].size() != 1){ printf("-1\n"); return 0; } for(int i = 2; i <= mmax; ++i) if(G[i].size() > (LL)G[i-1].size() * (m-1)){ printf("-1\n"); return 0; } printf("%d\n", n-1); for(int i = 0; i < mmax; ++i){ bool ok = true; int p = 0; for(int j = 0; j < G[i].size() && ok; ++j){ int u = G[i][j]; for(int k = (i != 0); k < m && ok; ++k){ printf("%d %d\n", u, G[i+1][p++]); if(p == G[i+1].size()) ok = false; } } } return 0; }
CodeForces 404C Restore Graph (构造)
标签:clu gcd can debug freopen ring queue comm define
原文地址:http://www.cnblogs.com/dwtfukgv/p/7107248.html