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最短路径问题
Time Limit: 1000 ms Memory Limit: 65536 KiB
Problem Description
平面上有n个点(n<=100),每个点的坐标均在-10000~10000之间。其中的一些点之间有连线。若有连线,则表示可从一个点到达另一个点,即两点间有通路,通路的距离为两点间的直线距离。现在的任务是找出从一点到另一点之间的最短距离。
Input
第1行为整数n。
第2行到第n+1行(共n行),每行两个整数x和y,描述了一个点的坐标(以一个空格分隔)。
第n+2行为一个整数m,表示图中连线的个数。
此后的m行,每行描述一条连线,由两个整数i和j组成,表示第i个点和第j个点之间有连线。
最后一行:两个整数s和t,分别表示源点和目标点。
Output
仅1行,一个实数(保留两位小数),表示从s到t的最短路径长度。
Sample Input
5
0 0
2 0
2 2
0 2
3 1
5
1 2
1 3
1 4
2 5
3 5
1 5
Sample Output
3.41
Hint
Source
Floyed:
#include <bits/stdc++.h>
#define INF INT_MAX
using namespace std;
struct node
{
int x, y; // 存坐标
} a[105];
int n, m;
double gra[105][105];
void Floyed()
{
for (int k = 1; k <= n; k++)
{
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
if (gra[i][j] > gra[i][k] + gra[k][j])
gra[i][j] = gra[i][k] + gra[k][j];
}
}
}
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d %d", &a[i].x, &a[i].y);
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
if (i == j)
gra[i][j] = 0;
else
gra[i][j] = INF;
}
}
int u, v;
scanf("%d", &m);
for (int i = 0; i < m; i++)
{
scanf("%d %d", &u, &v);
double t = sqrt(pow((a[u].x - a[v].x), 2) + pow((a[u].y - a[v].y), 2)); // 计算两点之间的距离
if (gra[u][v] > t)
gra[u][v] = gra[v][u] = t;
}
Floyed();
scanf("%d %d", &u, &v);
printf("%.2lf\n", gra[u][v]);
return 0;
}
/***************************************************
User name: jk180602
Result: Accepted
Take time: 4ms
Take Memory: 248KB
Submit time: 2019-01-18 21:43:14
****************************************************/
Dijkstra:
#include <bits/stdc++.h>
#define INF INT_MAX
using namespace std;
struct node
{
int x, y;
} a[105];
int n, m;
double gra[105][105], dist[105];
int cat[105];
void Dijkstra(int k)
{
for (int i = 1; i <= n; i++)
dist[i] = gra[k][i];
dist[k] = 0;
cat[k] = 1;
for (int q = 1; q <= n; q++)
{
int min = INF;
int index = k;
for (int j = 1; j <= n; j++)
{
if (!cat[j] && dist[j] < min)
{
min = dist[j];
index = j;
}
}
cat[index] = 1;
for (int i = 1; i <= n; i++)
{
if (!cat[i] && gra[index][i] < INF && dist[i] > gra[index][i] + dist[index])
dist[i] = gra[index][i] + dist[index];
}
}
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
scanf("%d %d", &a[i].x, &a[i].y);
}
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
if (i == j)
gra[i][j] = 0;
else
gra[i][j] = INF;
}
}
int u, v;
scanf("%d", &m);
for (int i = 0; i < m; i++)
{
scanf("%d %d", &u, &v);
double t = sqrt(pow((a[u].x - a[v].x), 2) + pow((a[u].y - a[v].y), 2));
if (gra[u][v] > t)
gra[u][v] = gra[v][u] = t;
}
scanf("%d %d", &u, &v);
memset(cat, 0, sizeof(cat));
Dijkstra(u);
printf("%.2lf\n", dist[v]);
return 0;
}
/***************************************************
User name: jk180602
Result: Accepted
Take time: 0ms
Take Memory: 244KB
Submit time: 2019-01-18 22:17:43
****************************************************/