PAT Advanced 1013

Battle Over Cities

简单的并查集应用：

思路

先假设这个点不在图里面，然后构建并查集，算有几个连通分量，算完之后，连通分量数-1就能得到需要修建多少条路。

源码

#include<iostream>#include<fstream>#include<cstring>using namespace std;int tree[1010];//ifstream fin("fin.txt");//streambuf *old = cin.rdbuf(fin.rdbuf());int N, M, K;int ans[1010];struct node{    int f, r;};node edge[1010*1010];void init(){    for (int i = 1; i <= N; i++)    {        tree[i] = i;        ans[i] = 0;    }}int findroot(int x){    int r = x;    while (tree[r] != r)        r = tree[r];    int i = x, j;    while (i != r)    {        j = tree[i];        tree[i] = r;        i = j;    }    return r;}void merge(int a, int b){    int fx = findroot(a), fy = findroot(b);    if (fx != fy)    {        tree[fy] = fx;    }}void solve(){    int city;    int highway=0;    for (int i = 1; i <= K; i++)    {        cin >> city;        init();        for (int j = 1; j <= M; j++)        {            if (edge[j].f != city &&edge[j].r != city)            {                merge(edge[j].f, edge[j].r);            }        }        for (int j = 1; j <= N; j++)        {            if (j != city)                findroot(j);        }        for (int j = 1; j <= N; j++)        {            if (tree[j]==j&&j!=city)            {                highway++;            }        }        if (!highway)            cout << 0 << endl;        else            cout << highway - 1 << endl;        highway = 0;    }}void input(){    cin >> N >> M >> K;    int c1, c2;    init();    for (int i = 1; i <= M; i++)    {        cin >> c1 >> c2;        edge[i].f = c1;        edge[i].r = c2;    }    if (N)        solve();    else        cout << 0 << endl;}int main(){    input();    return 0;}

目录