UVA - 10816 Travel in Desert(枚举＋生成树+最短路)

题目大意：给出N个点，M条边，每条边有相应的长度和温度。
现在给出起点和终点，问从起点到终点的路径中，温度最高的最小值是多少，路径路径长度是多少，经过了哪些点

解题思路：这题有两种解法

第一种：二分最高温度，然后跑最短路
第二种:按照kruscal算法来，按温度升序排序，当起点和终点连通的时候，那个时候的温度就是这条路的最高温度了
接着将所有低于等于该温度的边连成一张图，然后跑最短路即可

第一种思路:

#include <cstdio>#include <cstring>#include <queue>#include <cmath>using namespace std;#define N 110#define M 20010#define esp 1e-6#define INF 0x3f3f3f3fstruct Edge{    int v, next;    int t, d;}E[M];int head[N];int n, m, tot, source, sink; int Max, Min;void AddEdge(int u, int v, int t, int d) {    E[tot].v = v; E[tot].t = t; E[tot].d = d; E[tot].next = head[u]; head[u] = tot++;    u = u ^ v; v = u ^ v; u = u ^ v;    E[tot].v = v; E[tot].t = t; E[tot].d = d; E[tot].next = head[u]; head[u] = tot++;}void init() {    scanf("%d%d", &source, &sink);    Max = 0; Min = INF;    memset(head, -1, sizeof(head));    tot = 0;    int u, v, tt, dd;    double t, d;    for (int i = 0; i < m; i++) {        scanf("%d%d%lf%lf", &u, &v, &t, &d);        tt = (int)(round(t * 10.0));        dd = (int)(round(d * 10.0));        Max = max(Max, tt);        Min = min(Min, tt);        AddEdge(u, v, tt, dd);    }}struct Node{    int u, d;    Node() {}    Node(int u, int d): u(u), d(d) {}    bool operator < (const Node &a) const {        return d > a.d;    }};int pre[N];int d[N];bool vis[N];bool dijkstra(int mid) {    memset(vis, 0, sizeof(vis));    memset(pre, 0,sizeof(pre));    pre[source] = -1;    priority_queue<Node> Q;    for (int i = 1; i <= n; i++)          d[i] = INF;    d[source] = 0;    Q.push(Node(source, 0));    while (!Q.empty()) {        Node t = Q.top();        Q.pop();        if (vis[t.u]) continue;        vis[t.u] = true;        for (int i = head[t.u]; ~i; i = E[i].next) {            int v = E[i].v;            if (d[v] > d[t.u] + E[i].d && E[i].t <= mid) {                d[v] = d[t.u] + E[i].d;                pre[v] = t.u;                Q.push(Node(v, d[v]));            }        }    }    return d[sink] != INF;}void out(int v) {    if (pre[v] < 0) {        printf("%d", v);        return ;    }    out(pre[v]);    printf(" %d", v);}void solve() {    int l = Min, r = Max, mid;    while (r > l) {        double mid = (l + r) / 2;        if (dijkstra(mid)) r = mid;        else l = mid +  1;    }    dijkstra(r);    out(sink);    printf("\n");    printf("%.1lf %.1lf\n", d[sink] / 10.0, r / 10.0);}int main() {    while (scanf("%d%d", &n, &m) != EOF) {        init();        solve();    }    return 0;}

第二种思路：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>using namespace std;#define N 110#define M 10010#define INF 0x3f3f3f3fstruct Edge{    int u, v, t, d;    Edge() {}    Edge(int u, int v, int t, int d): u(u), v(v), t(t), d(d) {}}edge[M];struct Edge2{    int v, d, next;}E[2*M];int n, m, s, t, tot, ans_tmp;int f[N], head[N], d[N], pre[N];bool vis[N];int cmp(const Edge &a, const Edge &b) {    if (a.t == b.t)        return a.d < b.d;    return a.t < b.t;}void AddEdge(int u, int v, int d) {    E[tot].v = v; E[tot].d = d; E[tot].next = head[u]; head[u] = tot++;    u = u ^ v; v = u ^ v; u = u ^ v;    E[tot].v = v; E[tot].d = d; E[tot].next = head[u]; head[u] = tot++;}int find(int x) {    return x == f[x] ? x : f[x] = find(f[x]);   }void kruscal() {    for (int i = 1; i <= n; i++)        f[i] = i;    sort(edge, edge + m, cmp);    int xx, yy;    ans_tmp = INF;    for (int i = 0; i < m; i++) {        if (edge[i].t > ans_tmp) break;        xx = find(edge[i].u);        yy = find(edge[i].v);        AddEdge(edge[i].u, edge[i].v, edge[i].d);        if (xx != yy) {            f[xx] = yy;            if (find(s) == find(t)) ans_tmp = edge[i].t;        }    }}void spfa() {    for (int i = 1; i <= n; i++)        d[i] = INF;    memset(pre, -1, sizeof(pre));    memset(vis, 0, sizeof(vis));    d[s] = 0;    queue<int> Q;    Q.push(s);    while (!Q.empty()) {        int u = Q.front();        Q.pop();        for (int i = head[u]; ~i; i = E[i].next) {            int v = E[i].v;            if (d[v] > d[u] + E[i].d) {                d[v] = d[u] + E[i].d;                pre[v] = u;                if (!vis[v]) {                    vis[v] = true;                    Q.push(v);                }            }        }        vis[u] = false;    }}void print(int u) {    if (pre[u] == -1) {        printf("%d", u);        return ;    }    print(pre[u]);    printf(" %d", u);}void solve() {    memset(head, -1, sizeof(head));    tot = 0;    kruscal();    spfa();    print(t);    printf("\n");    printf("%.1lf %.1lf\n", d[t] / 10.0, ans_tmp / 10.0);}void init() {    scanf("%d%d", &s, &t);    int u, v, tt, dd;    double t, d;    for (int i = 0; i < m; i++) {        scanf("%d%d%lf%lf", &u, &v, &t, &d);        tt = (int)(round(t * 10.0));        dd = (int)(round(d * 10.0));        edge[i] = Edge(u, v, tt, dd);       }}int main() {    while (scanf("%d%d", &n, &m) != EOF) {        init();        solve();    }    return 0;}