uva 10816 Travel in Desert（簡單的好題~兩種方法）

標籤：uva

題意：

給出 一個圖

點與點之間的路徑上有兩個權值 路徑長度和溫度

要求在所走路徑中的溫度的最大值最小的前提下 走最短路徑

解題思路1：

首先用 最小產生樹 的方法走出 最小瓶頸路 ，把在這期間用到的所有溫度小於 路徑上最大溫度 的邊存下來，作為接下來求最短路徑的圖；

在新產生的圖中求最短路徑即可；

code

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<queue>using namespace std;const int maxm = 10005;const int maxn = 105;struct Edge{    int u,v;    double dist,tm;    void read(){        scanf("%d%d%lf%lf",&u,&v,&tm,&dist);        u--;v--;    }    bool operator<(const Edge et)const{        if(tm != et.tm) return tm < et.tm;        else return dist < et.dist;    }}e[maxm];int n,m,s,t;int parent[maxn];vector<Edge> g[maxn];void init(){    scanf("%d%d",&s,&t);    s--;t--;    for(int i = 0; i < m; i++){        e[i].read();    }    for(int i = 0; i < n; i++){            g[i].clear();    }}int find(int x){    if(parent[x] == x) return x;    else return parent[x] = find(parent[x]);}const int INF = 0x3f3f3f3f;const int MAXNODE = 105;struct Edge2 {    int u, v;    double dist;    Edge2() {}    Edge2(int u, int v, double dist) {        this->u = u;        this->v = v;        this->dist = dist;    }};struct HeapNode {    double d;    int u;    HeapNode() {}    HeapNode(double d, int u) {        this->d = d;        this->u = u;    }    bool operator < (const HeapNode& c) const {        return d > c.d;    }};struct Dijkstra {    int n, m;    vector<Edge2> edges;    vector<int> g[MAXNODE];    bool done[MAXNODE];    double d[MAXNODE];    int p[MAXNODE];    void init(int tot) {        n = tot;        for (int i = 0; i < n; i++)            g[i].clear();        edges.clear();    }    void add_Edge(int u, int v, double dist) {        edges.push_back(Edge2(u, v, dist));        m = edges.size();        g[u].push_back(m - 1);    }    void print(int s, int e) {//shun xu        if (s == e) {            printf("%d", e + 1);            return;        }        print(s, edges[p[e]].u);        printf(" %d", e + 1);    }    void print2(int s, int e) {//ni xu        if (s == e) {            printf("%d", e + 1);            return;        }        printf("%d ", e + 1);        print2(s, edges[p[e]].u);    }    void dijkstra(int s) {        priority_queue<HeapNode> Q;        for (int i = 0; i < n; i++) d[i] = INF*1.0;        d[s] = 0.0;        memset(done, false, sizeof(done));        Q.push(HeapNode(0, s));        while (!Q.empty()) {            HeapNode x = Q.top(); Q.pop();            int u = x.u;            if (done[u]) continue;            done[u] = true;            for (int i = 0; i < g[u].size(); i++) {                Edge2& e = edges[g[u][i]];                if (d[e.v] > d[u] + e.dist) {                    d[e.v] = d[u] + e.dist;                    p[e.v] = g[u][i];                    Q.push(HeapNode(d[e.v], e.v));                }            }        }    }} graph;void solve(){//    printf("...\n");    double ans = 0;    sort(e,e+m);//    for(int i = 0; i < m; i++){//        printf("%.1lf %.1lf %d %d\n",e[i].dist,e[i].tm,e[i].u,e[i].v);//    }    for(int i = 0; i < n; i++) parent[i] = i;    double max_tm = 500.0;    graph.init(n);    for(int i = 0; i < m; i++){        if(e[i].tm > max_tm) break;        graph.add_Edge(e[i].u,e[i].v,e[i].dist);        graph.add_Edge(e[i].v,e[i].u,e[i].dist);        int pu = find(e[i].u);        int pv = find(e[i].v);        if(pu == pv) continue;        parent[pu] = pv;        if(find(s) == find(t)){            max_tm = e[i].tm;        }    }    graph.dijkstra(s);    graph.print(s,t);    printf("\n");    printf("%.1lf %.1lf\n",graph.d[t],max_tm);}int main(){    while(scanf("%d%d",&n,&m) != EOF){        init();        solve();    }    return 0;}

解題思路二：

把原圖存下來，然後二分溫度，再把所有小於溫度mid的邊拿出來構成一個新圖，然後繼續dijkstra求最短路，有成功和不成功兩種結果，找到能成功的最小溫度即可

code

#include <cstdio>#include <cstring>#include <vector>#include <queue>using namespace std;const int MAXNODE = 105;const int MAXEDGE = 20005;typedef double Type;const Type INF = 0x3f3f3f3f;struct Edge {    int u, v;    Type dist, d;    Edge() {}    Edge(int u, int v, Type dist, Type d = 0) {        this->u = u;        this->v = v;        this->dist = dist;        this->d = d;    }    void read() {        scanf("%d%d%lf%lf", &u, &v, &d, &dist);        u--; v--;    }};struct HeapNode {    Type d;    int u;    HeapNode() {}    HeapNode(Type d, int u) {        this->d = d;        this->u = u;    }    bool operator < (const HeapNode& c) const {        return d > c.d;    }};int n, m, s, t;struct Dijkstra {    int n, m;    Edge edges[MAXEDGE];    int first[MAXNODE];    int next[MAXEDGE];    bool done[MAXNODE];    Type d[MAXNODE];    int p[MAXNODE];    void init(int n) {        this->n = n;        memset(first, -1, sizeof(first));        m = 0;    }    void add_Edge(int u, int v, Type dist) {        edges[m] = Edge(u, v, dist);        next[m] = first[u];        first[u] = m++;    }    void add_Edge(Edge e) {        edges[m] = e;        next[m] = first[e.u];        first[e.u] = m++;    }    void print(int e) {//shun xu        if (p[e] == -1) {            printf("%d", e + 1);            return;        }        print(edges[p[e]].u);        printf(" %d", e + 1);    }    void print2(int e) {//ni xu        if (p[e] == -1) {            printf("%d\n", e + 1);            return;        }        printf("%d ", e + 1);        print2(edges[p[e]].u);    }    bool dijkstra(int s, int t) {        priority_queue<HeapNode> Q;        for (int i = 0; i < n; i++) d[i] = INF;        d[s] = 0;        p[s] = -1;        memset(done, false, sizeof(done));        Q.push(HeapNode(0, s));        while (!Q.empty()) {            HeapNode x = Q.top(); Q.pop();            int u = x.u;            if (u == t)                return true;            if (done[u]) continue;            done[u] = true;            for (int i = first[u]; i != -1; i = next[i]) {                Edge& e = edges[i];                if (d[e.v] > d[u] + e.dist) {                    d[e.v] = d[u] + e.dist;                    p[e.v] = i;                    Q.push(HeapNode(d[e.v], e.v));                }            }        }        return false;    }} gao;Edge e[MAXEDGE];bool judge(double mid) {    gao.init(n);    for (int i = 0; i < m; i++) {        if (e[i].d > mid) continue;        gao.add_Edge(e[i]);        gao.add_Edge(Edge(e[i].v, e[i].u, e[i].dist, e[i].d));    }    if (gao.dijkstra(s, t)) return true;    return false;}int main() {    while (~scanf("%d%d", &n, &m)) {        scanf("%d%d", &s, &t);        s--; t--;        for (int i = 0; i < m; i++)            e[i].read();        double l = 0, r = 50, mid;        while (r - l > 1e-8) {            mid = (l + r) / 2;            if (judge(mid)) r = mid;            else l = mid;        }        if (judge(r)) {            gao.print(t); printf("\n");            printf("%.1lf %.1lf\n", gao.d[t], mid);        }    }    return 0;}

二分在很多情況下都是很好用的一種方法~

uva 10816 Travel in Desert（簡單的好題~兩種方法）