UVA10986 - Sending email(Dijkstra)

UVA10986 - Sending email(Dijkstra)

题目链接

题目大意：给n个点，m条边，还有起点和终点，问起点到终点的最短距离，不可达unreachable。

解题思路：最短路问题，dijkstra算法。

代码：

#include <cstdio>
#include <queue>
#include <vector>
#include <string.h> 

using namespace std;
using std::make_pair;

typedef pair<int, int> pii;
priority_queue<pii, vector<pii>, greater<pii> >q;

const int INF = 0x3f3f3f3f;
const int maxn = 2e4;
const int maxm = 1e5 + 5;

int first[maxn];
int u[maxm], v[maxm], w[maxm], next[maxm];
int d[maxn];

void read_Graph (int n, int m) {

    for (int i = 0; i < n; i++)
        first[i] = -1;

    m *= 2;
    for (int i = 0; i < m; i++) {
        scanf ("%d%d%d", &u[i], &v[i], &w[i]);
        next[i] = first[u[i]];
        first[u[i]] = i;

        i++;
        u[i] = v[i - 1];
        v[i] = u[i - 1];
        w[i] = w[i - 1];
        next[i] = first[u[i]];
        first[u[i]] = i;
    }
}

int Dijkstra (int s, int t, int n) {

    for (int i = 0; i < n; i++)
        d[i] = (i == s) ? 0 : INF;

    q.push(make_pair(d[s],s));

    pii cur;
    while (!q.empty()) {

        cur = q.top();
        q.pop();

        if (cur.first != d[cur.second]) continue;

        for (int i = first[cur.second]; i != -1; i = next[i])
            if (d[v[i]] > d[u[i]] + w[i]) {
                d[v[i]] = d[u[i]] + w[i];
                q.push(make_pair(d[v[i]], v[i]));
            }
    }

    return d[t];
}

int main () {

    int T;
    scanf ("%d", &T);

    int n, m, s, t, cas = 0;
    while (T--) {

        scanf ("%d%d%d%d", &n, &m, &s, &t);

        read_Graph(n, m);
        int ans = Dijkstra(s, t, n);

        if (ans == INF)
            printf ("Case #%d: unreachable\n", ++cas);
        else printf ("Case #%d: %d\n", ++cas, ans);
    }
    return 0;
}