1013. Battle Over Cities

来源：互联网发布：控制射精知乎编辑：程序博客网时间：2024/05/18 05:57

题目链接：http://pat.zju.edu.cn/contests/pat-a-practise/1013

并查集

// 思绪太乱，没有想好就写// 另外，所用变量名字，大小写都出现，生出不少端倪// 图的并查集问题有些生疏// // 本题思路：// 1.去掉被占领城市所有的边，计算合并次数ans, //  则ans-1为结果（去掉连通占领城市的边）// 2.合并时，跳过被占领的城市#include <stdio.h>#define SIZE 1000+10#define INF -1int map[SIZE][SIZE], nmap[SIZE][SIZE];int n, m, K;int fa[SIZE];int getfa(int x){if(x == fa[x])return x;elsefa[x] = getfa(fa[x]);return fa[x];}void Init(){int i;for(i=1; i<=n; i++){for(int j=1; j<=n; j++){if(i != j){map[i][j] = INF;}else{map[i][j] = 0;}}}return ;}void Initfa(){int i, j;for(i=1; i<=n; i++){fa[i] = i;}for(j=1; j<=n; j++){int k;for(k=j; k<=n; k++){int a= getfa(j);int b= getfa(k);if(nmap[j][k] == 1){fa[b] = a;}}}// 连通的节点合并return ;}void newMap(int city){int i, j;for(i=1; i<=n; i++){for(j=1; j<=n; j++){if(i == city || j == city){nmap[i][j] = INF;}else{nmap[i][j] = map[i][j];}}}return ;}void Printmap(int map[SIZE][SIZE]){int i, j;for(i=1; i<=n; i++){for(j=1; j<=n; j++){printf("%2d ", map[i][j]);}printf("\n");}return ;}int main(){#ifdef ONLINE_JUDGE#elsefreopen("E:\\in.txt", "r", stdin);//freopen("E:\\out.txt", "w", stdout);#endifscanf("%d%d%d", &n, &m, &K);Init();int i, c1, c2;for(i=0; i<m ; i++){scanf("%d%d", &c1, &c2);map[c1][c2] = map[c2][c1] = 1;}//Printmap(map);while(K-->0){if(n <= 2){printf("0\n");continue;}int lc;//losted cityscanf("%d", &lc);newMap(lc);Initfa();//printf("\n");//Printmap(nmap);int ans=0, j;for(j=1; j<=n; j++){if(j == lc){continue;}int k;for(k=j; k<=n; k++){if(k ==lc){continue;}int a= getfa(j);int b= getfa(k);if(a != b){ans++;fa[b] = a; }}}printf("%d\n", ans);}return 0;}

0 0