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Output For each test case ,output the least groups that needs to form to achieve their goal.?
Sample Input3 3 1 2 2 3 1 3 4 2 1 2 3 4
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Sample Output1 2 Hint New ~~~ Notice: if there are no road connecting one town ,tony may forget about the town. In sample 1,tony and his friends just form one group,they can start at either town 1,2,or 3. In sample 2,tony and his friends must form two group.? ????????? 题目大意：其实题目意思就是给你一幅图，问最少用多少笔可以把整幅图划完，每条边只能经过一次，孤立点忽略不计。这个就是欧拉回路的问题，判断划几笔要看入度和出度的值，还有入度-出度的值，判断有多少个欧拉路。用并查集来判断连通性，度数判断欧拉回路。不多说，下面代码。 链接：http://acm.hdu.edu.cn/showproblem.php?pid=3018 代码：
#include <iostream> #include <stdio.h> #include <memory.h> #include <vector> using namespace std; int father[100005], deg[100005], odd[100005]; bool hash[100005]; vector<int> a; int n, m, sum; void init() { int i; a.clear(); memset(hash, false, sizeof(hash)); memset(deg, 0, sizeof(deg)); memset(odd, 0, sizeof(odd)); for(i = 1; i <= n; i++) father[i] = i; } int find(int x) { if(x != father[x]) { father[x] = find(father[x]); } return father[x]; } void Union(int x, int y) { father[x] = y; } int main() { int i, k, x, y, fx, fy; while(scanf("%d %d", &n, &m) != EOF) { init(); for(i = 1; i <= m; i++) { scanf("%d %d", &x, &y); deg[x]++; deg[y]++; fx = find(x); fy = find(y); if(fx != fy) Union(fx, fy); } for(i = 1; i <= n; i++) { k = find(i); if(!hash[k]) { a.push_back(k); //a保存着集合，集合就是图 hash[k] = true; } if(deg[i]%2 == 1) odd[k]++; //保存这个集合的奇数度数的个数 } sum = 0; for(i = 0; i < a.size(); i++) { k = a[i]; if(deg[k] == 0) continue; //孤立点 if(odd[k] == 0) sum++; //改集合是欧拉回路，有一条路 else sum += odd[k]/2; } printf("%d\n", sum); } return 0; }?
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