UVA10986 - Sending email(Dijkstra)

来源：互联网发布：搜索与回溯算法讲解编辑：程序博客网时间：2024/05/17 01:34

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;}

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