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
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 0x3f3f3f3f
int 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: Accepted Take time: 12ms Take Memory: 208KB Submit 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 0x3f3f3f3f
int 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 Answer Take time: 16ms Take Memory: 208KB Submit 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 9999999
int 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 Answer Take time: 12ms Take Memory: 192KB Submit time: 2017-02-17 18:41:44 ****************************************************/