【九度OJ】1035【Floyd算法】

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抽象成图。
代码：
package Test1;import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;import java.io.StreamTokenizer;public class Test30_1035 {public static void main(String[] args) throws IOException {StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));//  Floyd算法是一个经典的动态规划算法while (true) {st.nextToken();int n = (int) st.nval;st.nextToken();int m = (int) st.nval;if (n == 0 && m == 0)break;int dist[][] = new int[27][27]; // 1--->26for (int i = 1; i <= 26; i++){for (int j = 1; j <= 26; j++){dist[i][j] = 1000000;  //设为Integer.MAX_VALUE 相加时会溢出}dist[i][i]=0;}String str;int child, parent1, parent2;for (int i = 0; i < n; i++) {st.nextToken();str = st.sval;child = str.charAt(0) - 64; // 从1开始parent1 = str.charAt(1) - 64;parent2 = str.charAt(2) - 64;if(parent1<=26 && parent1>=1)dist[parent1][child]=1;  //有向图    父母指向孩子if(parent2<=26 && parent2>=1)dist[parent2][child]=1;}for(int k=1;k<=26;k++)for(int i=1;i<=26;i++)for(int j=1;j<=26;j++){if(dist[i][k]+dist[k][j]<dist[i][j]){dist[i][j]=dist[i][k]+dist[k][j];}}int people1, people2;for (int i = 0; i < m; i++) {st.nextToken();str = st.sval;people1 = str.charAt(0) - 64;people2 = str.charAt(1) - 64;if(dist[people1][people2]!=1000000){print(dist[people1][people2],1);}else if(dist[people2][people1]!=1000000){print(dist[people2][people1],0);}else{System.out.println("-");}}}}private static void print(int i,int flag) {if(i==1){if(flag==1)System.out.println("child");elseSystem.out.println("parent");}else if(i==2){if(flag==1)System.out.println("grandparent");elseSystem.out.println("grandchild");}else{int n=i-2;for(int j=0;j<n;j++)System.out.print("great-");if(flag==1)System.out.println("grandparent");elseSystem.out.println("grandchild");}}}
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