问题 B: DS_7.2 求解连通分量个数(by Yan)

问题 B: DS_7.2 求解连通分量个数(by Yan)

时间限制: 15 Sec  内存限制: 256 MB
提交: 1105  解决: 343
[提交] [状态] [讨论版] [命题人:zengyan]

题目描述

从键盘接收图的顶点集，关系集，创建无向图。
第一行依次输入图的顶点个数n，关系个数k，以空格隔开。顶点个数<=20
第二行依次输入顶点值，类型为字符。
接下去有k行，每行为两个字符 u 和 v，表示节点u 和 v 连通。格式为【uv】,中间不用空格间隔。
计算连通分量个数并输出。
输出一个整数，表示连通分量个数。、

输入：

6 7
ABCDEF
AB
AE
BC
CD
DA
DB
EC

输出
2

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct {
    int arcs[50][50];
    char vex[50];
    int vexnum;
    int arcnum;
} AdjMatrix, *Matrix;
typedef struct {
    char data;
    int chu;
    int ru;
} vac[20];
vac v;
int visited[20];
void calculate(Matrix G) {
    v->chu = 0;
    v->ru = 0;
    for (int i = 0; i < G->vexnum; ++i) {
        for (int j = 0; j < G->vexnum; ++j) {
            v[i].data = G->vex[i];
            if (G->arcs[i+1][j+1] == 1) {
                v[i].chu++;
            }
            if (G->arcs[j+1][i+1] == 1) {
                v[i].ru++;
            }
        }
    }

    for (int k = 0; k < G->vexnum; ++k) {
        visited[k] = v[k].data;
       // printf("%c %d %d %d\n",v[k].data,v[k].chu,v[k].ru,v[k].ru+v[k].chu);
    }
}
int located(Matrix G, char ch) {
    int t;
    for (int i = 0; i < G->vexnum; i++) {
        if (G->vex[i] == ch) {
            t = i;
            break;
        }
    }
    return t;
}

void func(Matrix Graph) {
    for (int i = 1; i <= Graph->arcnum; i++) {
        char temp1[2];
        scanf("%s", temp1);
        int x = located(Graph, temp1[0]);
        int y = located(Graph, temp1[1]);
        Graph->arcs[x + 1][y + 1] = 1;
    }
    return;
}

void creat(Matrix Graph) {
    scanf("%d %d", &Graph->vexnum, &Graph->arcnum);
    for (int i = 1; i <= Graph->vexnum; i++) {
        for (int j = 1; j <= Graph->vexnum; j++) {
            Graph->arcs[i][j] = 0;
        }
    }
    scanf("%s", Graph->vex);
    func(Graph);
}

//one is the first index in Graph
//zero is the first index in storage vocabulary;
void Dfs(Matrix G,char ch){
   // printf("%c",ch);
    int x = located(G, ch);
    visited[x+1] = 1;
    for (int i = 1; i <= G->vexnum; ++i) {
        if(G->arcs[x+1][i]==1&&visited[i]!=1){
            Dfs(G,G->vex[i-1]);
        }
    }
}
/*void Bfs(Matrix g,char ch){

    int vis[20]={0};
    int x = located(g, ch);
    vis[x+1] = 1;
    for (int i = 1; i <= g->vexnum; ++i) {
        for (int j = 1; j <= g->vexnum; ++j) {
            if (g->arcs[i][j]==1&&vis[j]!=1){
                printf("%c",g->vex[j-1]);
                x = located(g, g->vex[j-1]);
                vis[x+1] = 1;
            }
        }
    }
}*/
int main() {
    memset(visited,0, sizeof(int));
    Matrix graph = (Matrix) malloc(sizeof(AdjMatrix));
    creat(graph);
    calculate(graph);
    Dfs(graph,v[0].data);
    //printf("\n%c",v[0].data);
    //Bfs(graph,v[0].data);
    int count = 1;
    for(int i = 1; i <= graph->vexnum; i++){
        if(visited[i]==0){
            Dfs(graph,v[i-1].data);
            count++;
        }
    }
    printf("%d",count);
    return 0;
}