图结构练习——最短路径（Dijkstra算法）

think:
1注意重复边的覆盖
2注意map数组的初始化
3注意dist数组的初始化

sdut原题链接

图结构练习——最短路径
Time Limit: 1000MS Memory Limit: 65536KB

Problem Description
给定一个带权无向图，求节点1到节点n的最短路径。

Input
输入包含多组数据，格式如下。
第一行包括两个整数n m，代表节点个数和边的个数。(n<=100)
剩下m行每行3个正整数a b c，代表节点a和节点b之间有一条边，权值为c。

Output
每组输出占一行，仅输出从1到n的最短路径权值。（保证最短路径存在）

Example Input
3 2
1 2 1
1 3 1
1 0

Example Output
1
0

Hint

Author
赵利强

以下为accepted代码

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;#define INF  0x3f3f3f3fint n, m;int map[104][104], vis[104], dist[104];void Dijkstra(int v){    int i, j, k;    for(i = 1; i <= n; i++)//dist数组的初始化    {        dist[i] = map[v][i];        vis[i] = 0;    }    dist[v] = 0;    vis[v] = 1;    for(i = 0; i < n-1; i++)    {        int min = INF, u = v;        for(j = 1; j <= n; j++)//寻找未标记结点的最小值        {            if(vis[j] == 0 && dist[j] < min)            {                u = j;                min = dist[j];            }        }        vis[u] = 1;        for(k = 1; k <= n; k++)//更新最短路        {            if(vis[k] == 0 && map[u][k] < INF && dist[k] > dist[u] + map[u][k])            {                dist[k] = dist[u] + map[u][k];            }        }    }}int main(){    int i, j, a, b, c;    while(scanf("%d %d", &n, &m) != EOF)    {        memset(vis, 0, sizeof(vis));        for(i = 1; i <= n; i++)        {            for(j = 1; j <= n; j++)            {                if(i == j)                    map[i][j] = 0;                else                    map[i][j] = INF;            }        }        for(i = 0; i < m; i++)        {            scanf("%d %d %d", &a, &b, &c);            if(map[a][b] > c)///避免覆盖最短路                map[a][b] = map[b][a] = c;        }        if(m == 0)            printf("0\n");        else        {            Dijkstra(1);            printf("%d\n", dist[n]);        }    }    return 0;}/***************************************************User name: Result: AcceptedTake time: 12msTake Memory: 208KBSubmit time: 2017-02-17 19:30:41****************************************************/

以下为wrong answer代码——Dijkstra算法理解不扎实，导致变量位置使用错误（将u的位置写成了v）

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;#define INF  0x3f3f3f3fint n, m;int map[104][104], vis[104], dist[104];void Dijkstra(int v){    int i, j, k;    for(i = 1; i <= n; i++)//dist数组的初始化    {        dist[i] = map[v][i];        vis[i] = 0;    }    dist[v] = 0;    vis[v] = 1;    for(i = 0; i < n-1; i++)    {        int min = INF, u = v;        for(j = 1; j <= n; j++)//寻找未标记结点的最小值        {            if(vis[j] == 0 && dist[j] < min)            {                u = j;                min = dist[j];            }        }        vis[u] = 1;        for(k = 1; k <= n; k++)//更新最短路        {            if(vis[k] == 0 && map[v][k] < INF && dist[k] > dist[u] + map[u][v])            {                dist[k] = dist[u] + map[u][k];            }        }    }}int main(){    int i, j, a, b, c;    while(scanf("%d %d", &n, &m) != EOF)    {        memset(vis, 0, sizeof(vis));        for(i = 1; i <= n; i++)        {            for(j = 1; j <= n; j++)            {                if(i == j)                    map[i][j] = 0;                else                    map[i][j] = INF;            }        }        for(i = 0; i < m; i++)        {            scanf("%d %d %d", &a, &b, &c);            if(map[a][b] > c)///避免覆盖最短路                map[a][b] = map[b][a] = c;        }        if(m == 0)            printf("0\n");        else        {            Dijkstra(1);            printf("%d\n", dist[n]);        }    }    return 0;}/***************************************************User name: Result: Wrong AnswerTake time: 16msTake Memory: 208KBSubmit time: 2017-02-17 19:27:15****************************************************/

以下为wrong answer代码——
1 没有考虑到重复边的覆盖问题
2 map数组初始化错误

#include <stdio.h>#include <string.h>#define INF 9999999int n, m;int map[104][104], dist[10400], vis[10400];void Dijkstra(int v){    for(int i = 1; i <= n; i++)    {        dist[i] = map[v][i];        vis[i] = 0;    }    vis[v] = 1;    dist[v] = 0;    for(int i = 0; i < n-1; i++)    {        int min = INF, u = v;        for(int j = 1; j <= n; j++)//寻找未访问的结点中的最小值        {            if(vis[j] == 0 && dist[j] < min)            {                u = j;                min =dist[j];            }        }        vis[u] = 1;        for(int k = 1; k <= n; k++)//更新        {            if(vis[k] == 0 && map[u][k] < INF && dist[k] > dist[u] + map[u][k])            {                dist[k] = dist[u] + map[u][k];            }        }    }}int main(){    int a, b, c;    while(scanf("%d %d", &n, &m) != EOF)    {        memset(map, 0, sizeof(map));        memset(vis, 0, sizeof(vis));        for(int i = 0; i < m; i++)        {            scanf("%d %d %d", &a, &b, &c);            map[a][b] = c;        }        Dijkstra(1);        printf("%d\n", dist[n]);    }    return 0;}/***************************************************User name: Result: Wrong AnswerTake time: 12msTake Memory: 192KBSubmit time: 2017-02-17 18:41:44****************************************************/