[SDOI 2009] Elaxia的路线

[题目链接]

         https://www.lydsy.com/JudgeOnline/problem.php?id=1880

[算法]

        最短路 + 动态规划

        时间复杂度 ： O(NlogN)

[代码]

         

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1510
const int INF = 2e9;

struct edge
{
        int to , w , nxt;
} e[MAXN * MAXN];

int tot , n , m , sp1 , sp2 , X1 , X2 , Y1 , Y2;
int head[MAXN] , dist1[MAXN] , dist2[MAXN] , dist3[MAXN] , dist4[MAXN] , f[MAXN] , visited[MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u ,int v , int w)
{
        tot++;
        e[tot] = (edge){v , w , head[u]};
        head[u] = tot;
}
inline void dijkstra(int *dist , int s)
{
        priority_queue< pair<int , int> , vector< pair<int , int> > , greater< pair<int , int> > > q;
        for (int i = 1; i <= n; i++) 
        {
                dist[i] = INF;
                visited[i] = false;
        }
        dist[s] = 0;
        q.push(make_pair(0 , s));
        while (!q.empty())
        {
                int cur = q.top().second;
                q.pop();
                if (visited[cur]) continue;
                visited[cur] = true;
                for (int i = head[cur]; i; i = e[i].nxt)
                {
                        int v = e[i].to , w = e[i].w;
                        if (dist[cur] + w < dist[v])
                        {  
                                dist[v] = dist[cur] + w;
                                q.push(make_pair(dist[v] , v));
                        }        
                }        
        }        
}
inline int dp(int u)
{
        if (f[u] != -1) return f[u];
        if (u == Y1    || u == Y2) return f[u] = 0;
        f[u] = 0;
        for (int i = head[u]; i; i = e[i].nxt)
        {
                int    v = e[i].to , w = e[i].w;
                if (dist1[u] + w + dist3[v] == sp1 && dist2[u] + w + dist4[v] == sp2)
                        chkmax(f[u] , dp(v) + w);
        }    
        return f[u];
}

int main()
{
        
        read(n); read(m);
        read(X1); read(Y1);
        read(X2); read(Y2);
        for (int i = 1; i <= m; i++)
        {
                int u , v , w;
                read(u); read(v); read(w);
                addedge(u , v , w);
                addedge(v , u , w);        
        }
        dijkstra(dist1 , X1);
        dijkstra(dist2 , X2);
        dijkstra(dist3 , Y1);
        dijkstra(dist4 , Y2);
        sp1 = dist1[Y1] , sp2 = dist2[Y2];
        memset(f , 255 , sizeof(f));
        for (int i = 1; i <= n; i++) dp(i);
        int ans = 0;
        for (int i = 1; i <= n; i++) chkmax(ans , f[i]);
        dijkstra(dist1 , Y1);
        dijkstra(dist3 , X1);
        swap(X1 , Y1);
        memset(f , 255 , sizeof(f));
        for (int i = 1; i <= n; i++) dp(i);
        for (int i = 1; i <= n; i++) chkmax(ans , f[i]);
        printf("%d
" , ans);
        
        return 0;
    
}