Codevs 1079 回家最短路 spfa || dijkstra

最新推荐文章于 2018-09-16 17:11:34 发布

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最新推荐文章于 2018-09-16 17:11:34 发布

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本文介绍了解决Codevs1079回家最短路径问题的方法，使用ASCII码将字母转换为顶点，并通过SPFA和Dijkstra算法求解最短路径。

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Codevs 1079 回家最短路

直接利用 ASCII 码把字母转化为顶点跑最短路就可以了，
如果嫌我的方法浪费了部分空间可以令把所有的 ASCII 都-‘A’；

spfa

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#define MAX_V 200
#define MAX_E (10000+10)
using namespace std;

int E, tot = 0;
int first[MAX_V], nxt[MAX_E << 1], dis[MAX_V];
bool used[MAX_V];

struct edge{
    int from, to, cost;
}es[MAX_E << 1];

void build(int ff, int tt, int dd)
{
    es[++tot] = (edge){ff,tt,dd};
    nxt[tot] = first[ff];
    first[ff] = tot;
}

queue <int> q;
void spfa(int s)
{
    memset(dis,63,sizeof(dis));
    dis[s] = 0;
    q.push(s);
    used[s] = 1;
    while(q.size())
    {
        int x = q.front();
        q.pop();
        used[x] = 0;
        for(int i = first[x]; i != -1; i = nxt[i])
        {
            int v = es[i].to;
            if(dis[v] > dis[x] + es[i].cost)
            {
                dis[v] = dis[x] + es[i].cost;
                if(!used[v])
                {
                    q.push(v);
                    used[v] = 1;
                }
            }
        }
    }
}

int deal(char c)
{
    return (int)(c);
}

char s[2];
int main()
{
    cin >> E;
    memset(first,-1,sizeof(first));
    for(int i = 1; i <= E; i ++)
    {
        int f, t, d;
        scanf("%s", s); f = deal(s[0]);
        scanf("%s", s); t = deal(s[0]);
        scanf("%d", &d);
        build(f,t,d); build(t,f,d);
    }

    spfa(deal('Z'));

    int minn = 1e9, ans;
    for(int i = deal('A'); i < deal('Z'); i ++)
        if(dis[i] < minn)
        {
            minn = dis[i];
            ans = i;
        }

    cout << (char)ans << " " << minn << endl;
    return 0;
}

dijkstra

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#define MAX_V 200
#define MAX_E (10000+10)
using namespace std;

int E, tot = 0;
int first[MAX_V], nxt[MAX_E << 1], dis[MAX_V];
bool done[MAX_V];

struct edge{
    int from, to, cost;
}es[MAX_E << 1];

void build(int ff, int tt, int dd)
{
    es[++tot] = (edge){ff,tt,dd};
    nxt[tot] = first[ff];
    first[ff] = tot;
}

struct zt{
    int v, d;
    bool operator < (const zt b) const
    {
        return d > b.d;
    }
};

priority_queue <zt> q;
void dijkstra(int s)
{
    memset(dis,63,sizeof(dis));
    dis[s] = 0;
    q.push((zt){s,dis[s]});
    while(q.size())
    {
        int x = q.top().v;
        q.pop();
        if(done[x]) continue;
        done[x] = 1;
        for(int i = first[x]; i != -1; i = nxt[i])
        {
            int v = es[i].to;
            if(dis[v] > dis[x] + es[i].cost)
            {
                dis[v] = dis[x] + es[i].cost;
                q.push((zt){v,dis[v]});
            }
        }
    }
}

int deal(char c)
{
    return (int)(c);
}

char s[2];
int main()
{
    cin >> E;
    memset(first,-1,sizeof(first));
    for(int i = 1; i <= E; i ++)
    {
        int f, t, d;
        scanf("%s", s); f = deal(s[0]);
        scanf("%s", s); t = deal(s[0]);
        scanf("%d", &d);
        build(f,t,d); build(t,f,d);
    }

    dijkstra(deal('Z'));

    int minn = 1e9, ans;
    for(int i = deal('A'); i < deal('Z'); i ++)
        if(dis[i] < minn)
        {
            minn = dis[i];
            ans = i;
        }

    cout << (char)ans << " " << minn << endl;
    return 0;
}