為幫助大家能在6月18日的比賽中有一個更好的成績，我會將藍橋杯官網上的歷屆決賽題目的四類語言題解都發出來。希望能對大家的成績有所幫助。

今年的最大目標就是能為【一億技術人】創造更高的價值。

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資源限制

內存限制：256.0MB C/C++時間限制：1.0s Java時間限制：3.0s Python時間限制：5.0s

C++

#include <bits/stdc++.h> using namespace std;const int MAXN = 500;int mat[25][25]; int link[205][205]; int moves[2][2] = {{1,0},{0,1}}; int match[MAXN]; bool vis[MAXN]; int n,m,num,total = 0;;void floyd(){for(int k = 1;k<=total;k++){for(int i = 1;i<=total;i++){for(int j = 1;j<=total;j++){if(!link[i][j] && link[i][k] && link[k][j])link[i][j] = 1;}}} }bool found(int x){for(int v = 1;v<=total;v++){if(!vis[v] && link[x][v]){vis[v] = true;if(match[v] == 0||found(match[v])){match[v] = x;return true;}}}return false; }int main(){scanf("%d %d",&n,&m);memset(mat,0,sizeof(mat));memset(match,0,sizeof(match));memset(link,0,sizeof(link));string str;for(int i = 0;i<n;i++){cin>>str;for(int j = 0;j<m;j++){num = str[j] - '0';if(num)mat[i][j] = ++total;else mat[i][j] = 0;}}for(int i = 0;i<n;i++){for(int j = 0;j<m;j++){if(!mat[i][j])continue;for(int k = 0;k<2;k++){int dx = i + moves[k][0];int dy = j + moves[k][1];if(dx < 0 || dy <0 || dx >= n || dy >=m)continue;if(mat[dx][dy])link[mat[i][j]][mat[dx][dy]] = 1;}}}floyd();int res = 0;for(int i = 1;i<=total;i++){memset(vis,0,sizeof(vis));if(found(i))res++;}printf("%d\n",total - res);return 0; }

Java

import java.io.*; import java.util.*;public class Main{static boolean[][] conect;static boolean[] visited;static int[] friend;static int sum = 1;public static void main(String[] args) throws IOException, InterruptedException{BufferedReader br = new BufferedReader(new InputStreamReader(System.in));String[] sl = br.readLine().split(" +");int N = Integer.parseInt(sl[0]);int M = Integer.parseInt(sl[1]);int[][] list = new int[N][M];for (int i = 0; i < N; i++) {String s = br.readLine();for (int j = 0; j < M; j++) {if(Integer.parseInt(s.charAt(j)+"") == 1)list[i][j] = sum++;}}friend = new int[sum];conect = new boolean[sum][sum];for (int i = 0; i < N; i++) {for (int j = 0; j < M; j++) {if(list[i][j]==0)continue;if(i!=N-1 && list[i+1][j] > 0) {conect[list[i][j]][list[i+1][j]] = true;}if(j!=M-1 && list[i][j+1] > 0) {conect[list[i][j]][list[i][j+1]] = true;}}}for (int i = 1; i < sum; i++) {for (int j = 1; j < sum; j++) {if(!conect[i][j])continue;for (int j2 = 0; j2 < sum; j2++) {if(conect[j][j2] && i!=j2)conect[i][j2] = true;}}}int TT = 0;for (int i = 1; i < sum; i++) {visited = new boolean[sum];if(xyl(i))TT++;}System.out.println(sum-1-TT);}static boolean xyl(int i) {for (int j = 1; j < sum; j++) {if(conect[i][j] && !visited[j]) {visited[j] = true;if(friend[j]==0 || xyl(friend[j])) {friend[j] = i;return true;}}}return false;}}

Python

# 弗洛伊德 def floyd(n):for k in range(1, n + 1): # 當前考慮的中間節點for i in range(1, n + 1): # 遍歷圖中的每一個節點for j in range(1, n + 1):# 考慮該節點的所有可能的鄰接節點if new_graph[i][k] * new_graph[k][j] == 1:new_graph[i][j] = 1 # 匈牙利 def dfs(node, node_cot):for k in range(1, node_cot+1):if new_graph[node][k] == 1:if visited[k] == 0:visited[k] = 1if connected[k] == -1 or dfs(connected[k] ,node_cot):connected[k] = nodereturn Truereturn Falsedx = [0, 1] dy = [1, 0] N, M = list(map(int, input().split())) graph = [[0 for i in range(M)] for j in range(N)]node_cot = 0 for i in range(N):strr = input()for j in range(M):e = int(strr[j])if e == 1:node_cot += 1 graph[i][j] = [e, node_cot]else:graph[i][j] = [e, -1]new_graph = [[0 for i in range(node_cot + 1)] for j in range(node_cot + 1)] for i in range(N):for j in range(M):if graph[i][j][0] == 1:node_idx = graph[i][j][1]for x in range(2):new_i = i + dx[x]new_j = j + dy[x]if 0 <= new_i < N and 0 <= new_j < M and graph[new_i][new_j][0] == 1:new_graph[node_idx][graph[new_i][new_j][1]] = 1floyd(node_cot)cot = 0visited = [0 for i in range(node_cot + 1)] connected = [-1 for i in range(node_cot + 1)] for i in range(1, node_cot + 1):visited = [0 for i in range(node_cot + 1)]if dfs(i, node_cot):cot += 1 print(node_cot - cot)

總結

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