uva10986(最短路径)

题目的意思就是

首先给出有几组测试样例

每组样例先是 给 n(点的个数 ),m(边的个数) ,s (起点),t (终点)..

然后是每条边 ,以及权值..

求最短路..

但是因为点的数量很多,要用优化的dijkstra + 优先队列.刘汝佳书里的模板差别只有这题是无向图.

#include<stdio.h>
#include<string.h>
#include<queue>
#include<vector>
using std::make_pair;
using namespace std;
typedef pair<int ,int> pii;
const int N = 400005;
const int M = 400005;
const int INF = 0x3f3f3f3f;
int first
;
int vis
;
int u[M],v[M],w[M],nex[M];
int d
;
int n,m,s,t;
priority_queue<pii ,vector<pii> ,greater<pii> > q;
void build() {
scanf("%d%d%d%d",&n,&m,&s,&t) ;
for (int i = 0 ;i < n ; i++) {
first[i] = -1;
}
m *= 2;
for (int e = 0 ; e < m ;e++) {
scanf("%d%d%d",&u[e],&v[e],&w[e]);
nex[e] = first[u[e]];
first[u[e]] = e;
e++;
v[e] = u[e - 1],u[e] = v[e - 1],w[e] = w[e - 1];
nex[e] = first[u[e]];
first[u[e]] = e;
}
}
void dijkstra() {
for (int i = 0 ; i < n ;i++) {
d[i] = (i == s ? 0 : INF);
}
memset(vis , 0 ,sizeof(vis));
q.push(make_pair(d[s],s));
while(!q.empty()) {
pii f = q.top();
q.pop();
int x = f.second;
if(vis[x]) continue;
vis[x] = 1;
for (int e = first[x]; e != -1; e = nex[e]) {
if(d[v[e]] > d[x] + w[e]) {
d[v[e]] = d[x] + w[e];
q.push(make_pair(d[v[e]] , v[e]));
}
}
}

}
int main () {
int cas = 1;
int tt;
scanf("%d", &tt);
while (tt--) {
build();
dijkstra();
printf("Case #%d: ", cas++);
if (d[t] == INF)
printf("unreachable\n");
else
printf("%d\n", d[t]);
}
return 0;
}