HDU_1232_畅通工程

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HDU 1232

思路：最基础的并查集，直接写就可以，最后看看总共有几个集合，则输出总集合个数-1就是需要修的路数。

#include <iostream>#include <iomanip>#include <cstring>#include <cstdio>#include <cstdlib>#include <cmath>#include <set>#include <map>#include <list>#include <stack>#include <deque>#include <queue>#include <vector>#include <algorithm>#include <functional>#define debug(x) cout << "--------------> " << x << endlusing namespace std;const double PI = acos(-1.0);const double eps = 1e-10;const long long INF = 0x7fffffff;const long long MOD = 1000000007;const int MAXN = 1000 + 7;int n, m;int pre[MAXN];int findRootNode(int x){        int r = x;        while(pre[r] != r)                r = pre[r];        int i = x;        //路径压缩        while(i != r)        {                int j = pre[i];                pre[i] = r;                i = j;        }        return r;}void setMerge(int x, int y){        int root_x, root_y;        root_x = findRootNode(x);        root_y = findRootNode(y);        if(root_x < root_y)                pre[root_y] = root_x;        else                pre[root_x] = root_y;}int main(){     while(~scanf("%d", &n))     {             if(n == 0)                break;             scanf("%d", &m);             memset(pre, 0, sizeof(pre));             for(int i = 1; i <= n; ++i)                pre[i] = i;             int city_A, city_B;             for(int i = 1; i <= m; ++i)             {                scanf("%d%d", &city_A, &city_B);                setMerge(city_A, city_B);             }             int ans = -1;             for(int i = 1; i <= n; ++i)             {                     if(findRootNode(i) == i)                        ans++;             }             printf("%d\n", ans);     }     return 0;}

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