c.cpp

好用 很犀利 可以 很简洁 相当的通俗

#include<stdio.h>
#define Q 100

void color(int g[][Q],int,char vexs[]);

void main (){
	char vexs[Q];
	int g[Q][Q];
	int n,e;
	int i,j,k;

	/*int g[13][13] = 
	{ 
		{0,0,0,0,1,1,1,0,0,1,0,0,0}, 
		{0,0,0,0,0,1,1,1,0,1,1,0,0}, 
		{0,0,0,0,0,0,0,0,0,1,1,1,0}, 
		{0,0,0,0,0,0,0,0,0,0,0,0,0}, 
        {1,0,0,0,0,0,0,1,0,0,1,0,0},
		{1,1,0,0,0,0,1,0,0,0,0,1,0},
		{1,0,0,0,0,1,0,0,0,0,1,1,0},
		{0,1,0,0,1,0,0,0,0,0,0,1,0},
		{0,0,0,0,0,0,0,0,0,0,0,0,0},
		{0,1,1,0,0,0,0,0,0,0,0,0,0},
		{0,1,1,0,1,1,1,0,0,0,0,0,0},
		{0,1,0,0,0,1,1,1,0,0,0,0,0},
		{0,0,0,0,0,0,0,0,0,0,0,0,0},
	}; */

	printf("输入路口的通路数和矛盾数:\n");
	scanf("%d%d",&n,&e);

	printf("输入通路信息:\n");
	for(i=0;i<n;i++)
		scanf("%s",&vexs[i]);

	for(i=0;i<n;i++)
		for(j=0;j<n;j++)
			g[i][j]=0; 
	printf("输入矛盾的通路号:\n");
		for(k=0;k<e;k++){
			scanf("%d%d",&i,&j);
			g[i][j]=1;
			g[j][i]=1;
		}

	printf("该交通图的邻接矩阵为:\n");
		for(i=0;i<n;i++){
			for(j=0;j<n;j++)
				printf("%d  ",g[i][j]);
			printf("\n");
		}
		 color(g,n,vexs);
}
	

void color(int  g[][Q],int n,char vexs[])
{
	int i,j,t=0,c[Q]={0};
	for(i=0;i<n;i++)
		for(j=0;j<n;j++){
			if(g[i][j]==1){
				if(c[i]==c[j]){
				c[j]=c[i]+1;
					if(t<c[j]+1)
						t=c[j]+1;
				}
			}	
		}
	printf("共需要设置%d种灯进行管理,\n",t);

	for(i=0;i<t;i++){
			printf("在%d号灯亮时以下路可通行:\n",i);
			for(j=0;j<n;j++){
			if(c[j]==i)
				printf("%c\n",vexs[j]);
	}
			}
}