POJ 2387 Til the Cows Come Home
題目大意:
無向圖中,從點n到點1的最短路徑
具體思路:
dijkstra,考慮重邊
注意下標減1,建議下次還是不使用0號,便於理解
具體代碼:
#include<iostream>
#include<cstdio>
#include<stdlib.h>
#include<string>
using namespace std;
int t, n;
const int N = 1005;
const int INF = 1e5;
int edgs[N][N];
int visit[N];
void init()
{
memset(visit, 0, sizeof(visit));
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
edgs[i][j] = INF;
}
void dijkstra()
{
visit[n - 1] = 1;
for (int i = 0; i < n - 1; i++)
{
int min_edg = INF, min_point;
for (int j = 0; j < n - 1; j++)
{
//得到距離最小的點
if (!visit[j] && edgs[n-1][j]<min_edg) {
min_edg = edgs[n - 1][j];
min_point = j;
}
}
//標記
min_edg = INF;
visit[min_point] = 1;
for (int j = 0; j < n - 1; j++)
{
//鬆弛各邊
if (!visit[j] && edgs[n - 1][j] > edgs[n - 1][min_point] + edgs[min_point][j]) {
edgs[n - 1][j] = edgs[n - 1][min_point] + edgs[min_point][j];
edgs[j][n - 1] = edgs[n - 1][j];
}
}
}
}
int main()
{
while (cin >> t >> n)
{
init();
for (int i = 0; i < t; i++)
{
int t1, t2, len;
cin >> t1 >> t2 >> len;
if (len < edgs[t1 - 1][t2 - 1]) { //考慮重邊
edgs[t1 - 1][t2 - 1] = len;
edgs[t2 - 1][t1 - 1] = len;
}
}
dijkstra();
cout << edgs[n - 1][0] << endl;
}
return 0;
}