"bzoj1179" Apio2009-atm

www.lydsy.com/JudgeOnline/problem.php?id=1179 (Topic link)

Test instructions: Give a graph, each node has a bit of power. Mark some points to find a path that can repeat through an edge, making the total point right and maximum. Repetition of a point can not be repeated to calculate the right.

Solution
Today's exam question, Dijkstra unfortunate GI rotten.
　　Warning:dijkstra handle the longest road when there will be some bad situation, so don't use!!
Now that we can repeat some of the edges, once we pass through a certain ring, we can always run all the points on the ring, so it is obvious. First Tarjan the contraction point, so the whole diagram becomes a direction-free graph, run DP or SPFA the longest road can be.

Code:

bzoj1179#include<algorithm> #include <iostream> #include <cstdlib> #include <cstring># include<cstdio> #include <cmath> #include <queue> #define LL long long#define inf 2147483640#define Pi ACOs ( -1.0) #define FREE (a) freopen (a ".", "R", stdin), Freopen (a ". Out", "w", stdout); using namespace Std;const int maxn=    500010;struct data {int num,x;    friend BOOL Operator < (const data &AMP;X,CONST data &y) {return x.x<y.x; }};struct Edge {int to,next;} e[maxn<<1];struct E {int u,v;} Ee[maxn];int Dis[maxn],head[maxn],dfn[maxn],low[maxn],st[maxn],vis[maxn],pos[maxn],a[maxn],w[maxn],ll[maxn];int N,m,top,sum,cnt,s,ind,p;void link (int u,int v) {e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;}    void Tarjan (int x) {dfn[x]=low[x]=++ind;    Vis[x]=1;    St[++top]=x;            for (int i=head[x];i;i=e[i].next) {if (!vis[e[i].to]) {Tarjan (e[i].to);        Low[x]=min (low[x],low[e[i].to]); } else if (!pos[e[i].to])           Low[x]=min (low[x],dfn[e[i].to]);        } if (Dfn[x]==low[x]) {sum++;        Int J;            do {j=st[top--];        POS[J]=SUM;W[SUM]+=A[J];    }while (st[top+1]!=x);    }}void Dijkstra () {priority_queue<data> q;    for (int i=1;i<=sum;i++) Dis[i]=-inf;    Data y,x= (data) {S,w[s]};    Q.push (x);d is[s]=w[s];        while (Q.size ()) {x=q.top (); Q.pop ();        if (Vis[x.num]) continue;        Vis[x.num]=1; for (int i=head[x.num];i;i=e[i].next) if (dis[e[i].to]<dis[x.num]+w[e[i].to]) {dis[e[i].to]=                Y.X=DIS[X.NUM]+W[E[I].TO];                y.num=e[i].to;            Q.push (y);    }}}void SPFA () {queue<int> q;    for (int i=1;i<=sum;i++) Dis[i]=-inf;    Q.push (S);d is[s]=w[s];        while (Q.size ()) {int X=q.front (); Q.pop ();        vis[x]=0; for (int i=head[x];i;i=e[i].next) if (dis[e[i].to]<dis[x]+w[e[i].to]) {dis[e[i].to]=dis[x]+w [e[I].TO];            if (!vis[e[i].to]) Q.push (e[i].to);    }}}int Main () {scanf ("%d%d", &n,&m);        for (int i=1;i<=m;i++) {scanf ("%d%d", &AMP;EE[I].U,&AMP;EE[I].V);    Link (EE[I].U,EE[I].V);    } for (int i=1;i<=n;i++) scanf ("%d", &a[i]);    scanf ("%d%d", &s,&p); Tarjan (S);    S=pos[s];        for (int x,i=1;i<=p;i++) {scanf ("%d", &x);    Ll[pos[x]]=1;    } for (int i=1;i<=n;i++) vis[i]=head[i]=0;    for (int i=1;i<=m;i++) if (POS[EE[I].U]!=POS[EE[I].V]) Link (pos[ee[i].u],pos[ee[i].v]);   Dijkstra ();    No SPFA ();    int ans=0;    for (int i=1;i<=sum;i++) if (Ll[i]) Ans=max (ans,dis[i]);    printf ("%d", ans); return 0;}

　　

"bzoj1179" Apio2009-atm