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Re:题解In Reply To:题解 Posted by:jwzxgo at 2012-03-10 03:33:50 distinct ... > 释疑: > - 一个人可以同时给几个人发送消息 > - 结束状态必须是所有人都受到了消息 > > 思路: > 1. 通过Stockbrokers(SB?)i发消息,所有人同时收到所花时间 (可以是无穷哦~) > 2. 找出第一步所得结果中的值最小的那一个 > 3. 如果第二步所得结果为无穷,说明此图不可联通~ > > 提示,题中数字均从1开始,如果程序中从0开始,需要相应地改计算公式,以及输出~ > > 欢迎发邮件一起讨论~ > > 源代码着色版:http://ideone.com/a32yM > > #include <stdio.h> > int nb=0; > int sp[100][100][101]; // i j k > int min(int a, int b){ if(a<=b) return a; return b; } > int i, j, k; > > void output(void){ > > int start_point = 0, time_cost = 9999; > for(i=0; i<nb; i++){ > int max_value = 0; > for( j=0; j<nb; j++ ){ > if( j!=i && max_value < sp[i][j][nb] ){ max_value=sp[i][j][nb]; } > } > if( time_cost > max_value ) { > time_cost = max_value; start_point = i; > } > } > if( time_cost >= 9999 ){ printf("distinct\n"); } > printf("%d %d\n", start_point+1, time_cost); > } > void cal(void){ > for(k=1; k<nb+1; k++){ > for(i=0; i<nb; i++){ > for(j=0; j<nb; j++){ > sp[i][j][k] = min(sp[i][j][k-1], sp[i][k-1][k-1]+sp[k-1][j][k-1]); > //printf("%d ", sp[i][j][k]); > }//printf("\n"); > }//printf("\n\n"); > } > } > void read_data(void){ > for(i=0; i<nb; i++){ > int t=0; > scanf("%d", &t); > for(j=0; j<t; j++){ > int to=0; > scanf("%d", &to); > scanf("%d", &(sp[i][to-1][0]) ); > } > } > } > void init(void){ > for(i=0; i<nb; i++){ > for(j=0; j<nb; j++){ > sp[i][j][0] = 9999; > } > } > } > int main(void) { > while(1){ > scanf("%d", &nb); > if( nb == 0 ) break; > init(); > read_data(); > cal(); > output(); > } > return 0; > } Followed by:
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