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[2016-01-21][HDU][1878]1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 | #include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cctype>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>using namespace std;long long d[1020];int fa[1020];int fnd(int x){ return (x == fa[x])?x:(fa[x] = fnd(fa[x]));}void uni(int x,int y){ fa[fnd(x)] = fnd(y);}void ini(int n){ for(int i = 0;i < n;i++) fa[i] = i;}int main(){ int n; long long m; while(~scanf("%d",&n) && n){ scanf("%I64d",&m); memset(d,0,sizeof(d)); ini(n + 10); //读取边,并把对应的端点加入并查集 int a,b; for(long long i = 0;i < m;i++){ scanf("%d%d",&a,&b); d[a]++;d[b]++; uni(a,b); } //判断度数 int flg = 1; for(int i = 1;flg && i <= n;i++) if(d[i] & 1) flg = 0; //判断联通分支数 set <int> s; if(flg){ for(int i = 1;i <= n;i++){ int f = fnd(i); if(s.count(f) == 0) s.insert(f); } } printf("%d\n",flg && (s.size() == 1)); } return 0;} |
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原文地址：http://www.cnblogs.com/qhy285571052/p/5149715.html
