NOIP2009 best Trade (BFS)

The positive solution is Tarjan, I did not write

With two BFs, the first BFS is done on the back of the original image, starting with N and finding all the points that can be reached from N.

The second BFS starts from the starting point, saving the minimum value mp[i] on each point to the N-point path.

Finally, the maximum value of w[i]-mp[i] can be calculated by traversing it again.

#include <cstdio> #include <iostream> #include <queue> #include <cstring> #define MAXN    100005using namespace std;struct t{int v; int next;}    Edge[500005],edge2[500005];int cnt,cnt2;int head[maxn],head2[maxn];void add_edge (int u,int v) {edge[cnt].v = v;    Edge[cnt].next = Head[u]; Head[u] = cnt++;}    void Add_edge2 (int u,int v)//save inverse diagram {edge2[cnt2].v = v;    Edge2[cnt2].next = Head2[u]; Head2[u] = cnt2++;} int Mp[maxn],w[maxn];bool able[maxn],vis[maxn];queue<int> myque;int n,m;void bfs1 () {for (int i = 1; I <= N; i+        +)//vis[i] Indicates whether I is in the queue, which avoids the effect of the ring in the diagram {Myque.push (i);    Vis[i] = 1;        } while (!myque.empty ()) {int u = myque.front ();        Vis[u] = 0;        Myque.pop ();            for (int i = head[u]; i =-1; i = edge[i].next) {int v = EDGE[I].V;                if (Mp[v] > Mp[u]) {Mp[v] = Mp[u];           if (!vis[v]) {vis[v] = 1;         Myque.push (v);    }}}}}void bfs2 () {Myque.push (n);    Able[n] = 1;        while (!myque.empty ()) {int u = myque.front ();        Myque.pop ();            for (int i = head2[u]; i =-1; i = edge2[i].next) {int v = EDGE2[I].V;                if (!able[v]) {able[v] = 1;            Myque.push (v);    }}}}int Main () {memset (head,-1,sizeof head);    memset (head2,-1,sizeof head2);    scanf ("%d%d", &n,&m);        for (int i = 1; I <= n; i++) {scanf ("%d", &w[i]);    Mp[i] = W[i];        } for (int i = 1; I <= m; i++) {int x, y, Z;        scanf ("%d%d%d", &x,&y,&z);        Add_edge (x, y);        Add_edge2 (Y,X);            if (z = = 2) {Add_edge (y,x);        Add_edge2 (x, y);    }} bfs2 ();//reverse, find can reach point bfs1 ();//forward, find I to n path maximum value int ans = 0; for (int i = 1; I <= n; i++) if (Able[i]) ans =Max (Ans,w[i]-mp[i]);    printf ("%d\n", ans); return 0;}

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NOIP2009 best Trade (BFS)