PAT 1013. Battle Over Cities

</pre></h1><h1 style="margin:0px 0px 0.5em; padding:0px; border:0px; font-family:'Droid Sans',Verdana,'Microsoft YaHei',Tahoma,sans-serif; font-size:3em; font-weight:normal; line-height:1em; vertical-align:baseline; color:rgb(17,17,17); background-color:rgb(250,250,250)"><pre style="margin-top:1.5em; margin-bottom:1.5em; padding:0px; border:0px; font-family:'Droid Sans Mono',Consolas,'Courier New',monospace; vertical-align:baseline; overflow:auto"><span style="font-size:18px"><span style="line-height:27px"></span></span><pre code_snippet_id="1914188" snippet_file_name="blog_20161006_2_3434556" name="code" class="java"><pre code_snippet_id="1914188" snippet_file_name="blog_20161006_1_4371864" name="code" class="java" style="color: rgb(17, 17, 17); font-size: 36px; line-height: 36px;">题目的意思就是求有多少个联通分量，首先想到的是用DFS，于是用Java写了一下：

package p1013;import java.util.Scanner;/* * 可以更优化 */public class Copy_2_of_Main {static boolean[] marked;static int[] cc;static String[] connected;static int N;static Scanner sc;public static void main(String[] args) {sc = new Scanner(System.in);N = sc.nextInt();int M = sc.nextInt();int Q = sc.nextInt();//marked = new boolean[1+N];cc = new int[1+N];connected = new String[1+N];for(int i=0; i<M; i++) {int s = sc.nextInt();int t = sc.nextInt();if(connected[s] == null)connected[s] = String.valueOf(t);elseconnected[s] += String.valueOf(t);//connected[t][s] = true;}for(int i=0; i<Q; i++) {System.out.println(CC());}}private static int CC() {marked = new boolean[1+N];marked[sc.nextInt()] = true;int cnt = 0;for(int i=1; i<=N; i++) {if(!marked[i]) {dfs(i);cnt ++;}}return cnt - 1;}private static void dfs(int start) {marked[start] = true;for(int i=1; i<=N; i++) {if(!marked[i] && (connected[start] != null || connected[i] != null)) {if(connected[start].indexOf(i+"") != -1 || connected[i].indexOf(start+"") != -1)dfs(i);}}}}

然而最后一个用例超时了，考虑到每次查一次都要DFS遍历整个图，其中很多都是重复的，所以改进了一下：

package p1013;import java.util.Scanner;/* * 可以更优化 */public class Main {static boolean[] marked;static int[] cc;static boolean[][] connected;static int N;static int s, sid;static int cnt = 0;public static void main(String[] args) {Scanner sc = new Scanner(System.in);N = sc.nextInt();int M = sc.nextInt();int Q = sc.nextInt();//marked = new boolean[1+N];cc = new int[1+N];connected = new boolean[1+N][1+N];for(int i=0; i<M; i++) {int s = sc.nextInt();int t = sc.nextInt();connected[s][t] = true;connected[t][s] = true;}int all = CC();for(int i=0; i<Q; i++) {s = sc.nextInt();sid = cc[s];System.out.println(all - 2 + CC2());}}private static int CC2() {cnt = 0;marked = new boolean[1+N];for(int i=1; i<=N; i++) {if(i != s && cc[i] == sid && !marked[i]) {dfs2(i);cnt ++;}}return cnt;}private static void dfs2(int start) {marked[start] = true;for(int i=1; i<=N; i++) {if(i != s && cc[i] == sid && !marked[i] && connected[start][i]) {dfs(i);}}}private static int CC() {cnt = 0;marked = new boolean[1+N];for(int i=1; i<=N; i++) {if(!marked[i]) {dfs(i);cnt ++;}}return cnt;}private static void dfs(int start) {marked[start] = true;cc[start] = cnt;for(int i=1; i<=N; i++) {if(!marked[i] && connected[start][i]) {dfs(i);}}}}

然而还是最后一个用例不通过，然后想到可能是Java运行本来就慢，于是换C++，艰难的照着网上的答案和自己的算法思路一样的敲了一遍，全部AC，而且时间比Java不是一般的少（最后一个用例也是特别变态）：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define MAX 1000// 会初始化？int n, m, k;bool marked[MAX];bool connected[MAX][MAX];void dfs(int t) {    marked[t] = true;    for(int i=1; i<=n; i++) {        if(!marked[i] && connected[t][i]) {            dfs(i);        }    }}int main(){    scanf("%d%d%d", &n, &m, &k);    int a, b;    for(int i=0; i<m; i++) {        scanf("%d%d", &a, &b);        connected[a][b] = true;        connected[b][a] = true;    }    int c;    while(k --) {        int count = 0;        scanf("%d", &c);        memset(marked, false, sizeof(marked));        marked[c] = true;        for(int i=1; i<=n; i++){            if(!marked[i]) {                dfs(i);                count++;            }        }        printf("%d\n", count-1);    }    return 0;}

测试点
测试点结果用时(ms)内存(kB)得分/满分0答案正确33847/71答案正确42568/82答案正确52443/33答案正确514083/34答案正确13714084/4

Java的什么Scanner啊这些好像都比scanf慢多了。。。。

网上说还有一种并查集的解法，到时候再试一下，现在是在是无力吐槽这个PAT了