POJ 1511 Invitation Cards （最短路的两种方法spfa， Dij）

题意：

给定n个点， m条路， 求1到 2 ~n的最短路之和加上2~n到1的最短路之和

分析：

裸最短路， 求其他点到源点的距离只需要把边方向再从源点求一次即可

spfa代码

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
#define rep(i,a,b) for(int i = a; i < b; i++)
#define _rep(i,a,b) for(int i = a; i <= b; i++)
#define mem(a,n) memset(a,n,sizeof(a))
#define fre(a) freopen(a,"r", stdin);
typedef long long LL;
using namespace std;
const LL inf = 1e12 + ;
const int maxn = 1e6 + ;
inline void read(int &x)
{
    int k=;
    char f=;
    char c=getchar();
    while(c>''||c<'')
        if(c=='-')
        {
            f=-;
            c=getchar();
        }
    while(c<=''&&c>='')
    {
        k=k*+c-'';
        c=getchar();
    }
    x =  k*f;
}
struct edge
{
    LL to, d;
    edge(LL _to, LL _d):to(_to), d(_d) {}
};
struct
{
    int to, next, d;
} node[][maxn];
int head[][maxn];
int cnt;
int n, m;
LL ans;
LL dis[maxn];
bool vis[maxn];
void dij(int st, int index)
{

    fill(dis, dis+n+, inf);
    mem(vis,);
    dis[st] = ;
    vis[st] = ;//记得入队标记第一个点
    queue<int> q;
    q.push();
    while(!q.empty())
    {
        int u = q.front();
        for(int i = head[index][u]; i != -; i = node[index][i].next)
        {
            int v = node[index][i].to;
            if(!vis[v] && dis[u] + node[index][i].d < dis[v])
            {
                dis[v] = dis[u] + node[index][i].d;
                if(!vis[v])
                    q.push(v);//松弛后如果没有在队中就入队
            }
        }
        q.pop();
        vis[u] = ;
    }
    LL sum = ;

    _rep(i,,n) ans += dis[i];
}
int main()
{
    int T;
    read(T);
    while(T--)
    {
        cnt = ;
        ans = ;
        mem(head[],-);
        mem(head[],-);
        read(n),  read(m);
        rep(i,,m)
        {
            int u, v, d;
            read(u) , read(v) , read(d) ,
            node[][cnt].to = v;
            node[][cnt].d = d;
            node[][cnt].next = head[][u];
            head[][u] = cnt;

            node[][cnt].to = u;
            node[][cnt].d = d;
            node[][cnt].next = head[][v];
            head[][v] = cnt++;
        }
        dij(,);
        dij(,);
        printf("%lld\n", ans);
    }
    return ;
}

dijkstra代码：

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
#define rep(i,a,b) for(int i = a; i < b; i++)
#define _rep(i,a,b) for(int i = a; i <= b; i++)
#define mem(a,n) memset(a,n,sizeof(a))
#define fre(a) freopen(a,"r", stdin);
typedef long long LL;
using namespace std;
const LL inf = 1e12 + ;
const int maxn = 1e6 + ;
inline void read(int &x)
{
    int k=;
    char f=;
    char c=getchar();
    while(c>''||c<'')
        if(c=='-')
        {
            f=-;
            c=getchar();
        }
    while(c<=''&&c>='')
    {
        k=k*+c-'';
        c=getchar();
    }
    x =  k*f;
}
struct edge
{
    LL to, d;
    edge(LL _to, LL _d):to(_to), d(_d) {}
};
struct
{
    int to, next, d;
} node[][maxn];
int head[][maxn];
int cnt;
int n, m;
LL ans;
LL dis[maxn];
bool vis[maxn];
void dij(int st, int index)
{

    fill(dis, dis+n+, inf);
    mem(vis,);
    dis[st] = ;//dij不需要标记第一个点
    priority_queue<pair<int,int>, vector< pair<int,int> >, greater<pair<int, int> > > q;//用pair的时候要记得优先队列如果不加greater<pair<int, int>>  是按从大到小排列的
    q.push(make_pair(,));
    while(!q.empty())
    {
        int u = q.top().second;
        q.pop();
        if(vis[u]) continue;
        vis[u] = ;
        for(int i = head[index][u]; i != -; i = node[index][i].next)
        {
            int v = node[index][i].to;
            if(!vis[v] && dis[u] + node[index][i].d < dis[v])
            {
                dis[v] = dis[u] + node[index][i].d;
                q.push(make_pair(dis[v],v));
            }
        }
    }
    LL sum = ;

    _rep(i,,n) ans += dis[i];
}
int main()
{
    int T;
    read(T);
    while(T--)
    {
        cnt = ;
        ans = ;
        mem(head[],-);
        mem(head[],-);
        read(n),  read(m);
        rep(i,,m)
        {
            int u, v, d;
            read(u) , read(v) , read(d) ,
            node[][cnt].to = v;
            node[][cnt].d = d;
            node[][cnt].next = head[][u];
            head[][u] = cnt;

            node[][cnt].to = u;
            node[][cnt].d = d;
            node[][cnt].next = head[][v];
            head[][v] = cnt++;
        }
        dij(,);
        dij(,);
        printf("%lld\n", ans);
    }
    return ;
}