D - Silver Cow Party J - Invitation Cards 最短路

http://poj.org/problem?id=3268

题目思路：

直接进行暴力，就是先求出举行party的地方到每一个地方的最短路，然后再求以每一个点为源点跑的最短路。

还有一种方法会快很多，就是跑两次最短路，一次正向的，另外一次反向的，因为都是只要求每一个位置到源点的最短距离，

所以这样子写就会快很多。

这个想法看完这个题目在脑袋里一闪而过没有仔细想，后来发现可以直接暴力，就忘记了，结果后面有一个数据大的就不行了。。。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#include <map>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 2e5 + ;
int d[maxn], dis[maxn], n, m;
struct node
{
    int from, to, dist;
    node(int from=,int to=,int dist=):from(from),to(to),dist(dist){}
};

struct heapnode
{
    int u, d;
    heapnode(int u=,int d=):u(u),d(d){}
    bool operator<(const heapnode&a)const
    {
        return a.d < d;
    }
};
vector<node>vec[maxn];
bool vis[maxn];
void dij(int s)
{
    priority_queue<heapnode>que;
    for (int i = ; i <= n; i++) d[i] = inf;
    d[s] = ;
    memset(vis, , sizeof(vis));
    que.push(heapnode(s, ));
    while(!que.empty())
    {
        heapnode x = que.top(); que.pop();
        int u = x.u;
        if (vis[u]) continue;
        vis[u] = ;
        for(int i=;i<vec[u].size();i++)
        {
            node e = vec[u][i];
            if(d[e.to]>d[u]+e.dist)
            {
                d[e.to] = d[u] + e.dist;
                que.push(heapnode(e.to, d[e.to]));
            }
        }
    }
}

int main()
{
    int k;
    scanf("%d%d%d", &n, &m, &k);
    for(int i=;i<=m;i++)
    {
        int x, y, c;
        scanf("%d%d%d", &x, &y, &c);
        vec[x].push_back(node(x, y, c));
    }
    int ans = ;
    dij(k);
    for (int i = ; i <= n; i++) dis[i] = d[i];
    for(int i=;i<=n;i++)
    {
        if (i == k) continue;
        dij(i);
    //    printf("dis[%d]=%d d[%d]=%d\n", i, dis[i], k, d[k]);
        ans = max(ans, dis[i] + d[k]);
    }
    printf("%d\n", ans);
    return ;
}

http://poj.org/problem?id=1511

这个和上面的题目一个意思

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#include <map>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 1e6 + ;
ll d[maxn], dis[maxn];
int n, m;
struct node
{
    int from, to;
    ll dist;
    node(int from = , int to = , ll dist = ) :from(from), to(to), dist(dist) {}
};

struct heapnode
{
    int u;
    ll d;
    heapnode(int u = , ll d = ) :u(u), d(d) {}
    bool operator<(const heapnode&a)const
    {
        return a.d < d;
    }
};
vector<node>vec[maxn];
bool vis[maxn];
void dij(int s)
{
    priority_queue<heapnode>que;
    for (int i = ; i <= n; i++) d[i] = inf;
    d[s] = ;
    memset(vis, , sizeof(vis));
    que.push(heapnode(s, ));
    while (!que.empty())
    {
        heapnode x = que.top(); que.pop();
        int u = x.u;
        if (vis[u]) continue;
        vis[u] = ;
        for (int i = ; i < vec[u].size(); i++)
        {
            node e = vec[u][i];
            if (d[e.to] > d[u] + e.dist)
            {
                d[e.to] = d[u] + e.dist;
                que.push(heapnode(e.to, d[e.to]));
            }
        }
    }
}
int a[maxn], b[maxn];
ll c[maxn];
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        for (int i = ; i <= n; i++) vec[i].clear();
        for (int i = ; i <= m; i++)
        {
            scanf("%d%d%lld", &a[i], &b[i], &c[i]);
            vec[a[i]].push_back(node(a[i],b[i],c[i]));
        }
        dij();
        for (int i = ; i <= n; i++)
        {
            dis[i] = d[i];
            vec[i].clear();
        }
        for(int i=;i<=m;i++)
        {
            vec[b[i]].push_back(node(b[i], a[i], c[i]));
        }
        dij();
        ll ans = ;
        for(int i=;i<=n;i++)
        {
            ans += d[i] + dis[i];
        }
        printf("%lld\n", ans);
    }
    return ;
}