暢通工程續
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 41691 Accepted Submission(s): 15410
現在,已知起點和終點,請你計算出要從起點到終點,最短需要行走多少距離。
每組數據第一行包含兩個正整數N和M(0<N<200,0<M<1000),分別代表現有城鎮的數目和已修建的道路的數目。城鎮分別以0~N-1編號。
接下來是M行道路信息。每一行有三個整數A,B,X(0<=A,B<N,A!=B,0<X<10000),表示城鎮A和城鎮B之間有一條長度爲X的雙向道路。
再接下一行有兩個整數S,T(0<=S,T<N),分別代表起點和終點。
#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
const int MAXN = 1005;
int dist[MAXN], len[MAXN][MAXN], vt[MAXN];
int n, m, s, t, a, b, d;
void Dijkstra(int root, int finish)
{
memset(vt, 0, sizeof(vt));
for(int i = 0;i < n; ++i)
{
dist[i] = 0x7FFFFFFF;
}
dist[root] = 0;
vt[root] = 1;
for(int i = 1;i <= n; ++i)
{
int minx = 0x7FFFFFFF;
for(int j = 0;j < n; ++j)
{
if(vt[j] == 1) continue;
if(len[root][j] == 0x7FFFFFFF) continue;
if(dist[j] > dist[root] + len[root][j])
{
dist[j] = dist[root] + len[root][j];
}
}
for(int j = 0;j < n; ++j)
{
if(vt[j] == 0 && dist[j] < minx)
{
root = j;
minx = dist[j];
}
}
vt[root] = 1;
}
if(dist[finish] == 0x7FFFFFFF) cout << -1 <<endl;
else cout << dist[finish] << endl;
}
int main()
{
while(scanf("%d %d", &n, &m) != EOF)
{
for(int i = 0;i < n; ++i)
for(int j = 0;j < n; ++j)
len[i][j] = 0x7FFFFFFF;
for(int i = 1;i <= m; ++i)
{
scanf("%d %d %d", &a, &b, &d);
if(d < len[a][b])
{
len[a][b] = len[b][a] = d;
}
}
scanf("%d %d", &s, &t);
Dijkstra(s, t);
}
return 0;
}