Bessie Come Home

题意：给定点、P条边、P条边的长度、终点、若干起点，求这若干起点到终点最短路径的最小值。（注意：两个点之间可能有多条边）

解题思路：

1. 读入输入，处理可以得到一张表table[i][j] = k，代表i点到j点有一条长为k的边（因为两个点之间可能有多条边，所以在处理过程中只保存较小的边）。并且可以得到起点和终点
2. 以终点为source进行Dijkstra，可以得到distance[i] = j，代表从source到节点i的最短路径。
3. 遍历distance[i] = j，找到其中i为起点情况下最小的j，那么就得到了最终的结果

代码：

/*ID: zc.rene1LANG: CPROG: comehome */#include<stdio.h>#include<stdlib.h>#include<string.h>#define MAX 20000000int P;int table[52][52];int visited[52];int cows_in[52];void GetInput(FILE *fin){    int i, j, distance;    char c[2];    int b[2];    for (i=0; i<52; i++)    {for (j=0; j<52; j++){    table[i][j] = MAX;}    }    for (i=0; i<52; i++)    {cows_in[i] = 0;visited[i] = 1;    }    fscanf(fin, "%d\n", &P);    for (i=0; i<P; i++)    {fscanf(fin, "%c %c %d\n", &c[0], &c[1], &distance);for (j=0; j<2; j++){    if (c[j] <= 'Z')    {b[j] = c[j] - 'A';if (c[j] != 'Z'){    cows_in[b[j]] = 1;}    }    else    {b[j] = c[j] - 'a' + 26;    }    visited[b[j]] = 0;}if (distance < table[b[0]][b[1]]){    table[b[0]][b[1]] = distance;    table[b[1]][b[0]] = distance;}    }}int FirstLeastUnvisted(int *distance){    int i, ret = 52, min = MAX;    for (i=0; i<52; i++)    {if (!visited[i]){    if (distance[i] < min)    {min = distance[i];ret = i;    }}    }    return ret;}void Dijkstra(int source, FILE *fout){    int distance[52];    int i, j, min_i, min = MAX;    for (i=0; i<52; i++)    {distance[i] = MAX;    }    distance[source] = 0;    while ((i = FirstLeastUnvisted(distance)) != 52)    {visited[i] = 1;for (j=0; j<52; j++){    if (table[i][j] != MAX)    {if ((distance[i] + table[i][j]) < distance[j]){    distance[j] = distance[i] + table[i][j];}    }}    }        for (i=0; i<52; i++)    {if (cows_in[i] == 1){    if (distance[i] < min)    {min = distance[i];min_i = i;    }}    }    fprintf(fout, "%c %d\n", min_i + 'A', min);}int main(void){    FILE *fin, *fout;    fin = fopen("comehome.in", "r");    fout = fopen("comehome.out", "w");    /*get input*/    GetInput(fin);    /*Dijkstra*/    Dijkstra(25, fout);    return 0;}