HDU 3986 Harry Potter and the Final Battle(Dijkstra)

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http://acm.hdu.edu.cn/showproblem.php?pid=3986

题意:

给你一个无向图,现在在删除图中任意一条边的情况下,要你求从1号点到N号点的最短距离的所有可能值中的最大值.

分析:

明显删除的这条边一定要在1到N的最短路径上才是对结果有影响的.所以我们只需要标记所有存在最短路径上的边,然后一次枚举不走该边的情况下计算最短路径值即可.

AC代码:

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#define INF 1e9using namespace std;const int maxn =1000+10;const int maxm=50000*2+10;struct Edge{    int from,to,dist;    Edge(){}    Edge(int f,int t,int d):from(f),to(t),dist(d){}};struct HeapNode{    int d,u;    HeapNode(int d,int u):d(d),u(u){}    bool operator<(const HeapNode &rhs)const    {        return d>rhs.d;    }};struct Dijkstra{    int n,m;    Edge edges[maxm];    int head[maxn],next[maxm];    int d[maxn];    bool done[maxn];    int p[maxn];    void init(int n)    {        this->n=n;        m=0;        memset(head,-1,sizeof(head));    }    void AddEdge(int from,int to,int dist)    {        edges[m]=Edge(from,to,dist);        next[m]=head[from];        head[from] = m++;    }    int dijkstra(int edge_id)//不走edge_id这条边    {        priority_queue<HeapNode> Q;        for(int i=0;i<n;i++) d[i]= i==0?0:INF;        memset(done,0,sizeof(done));        Q.push(HeapNode(d[0],0));        while(!Q.empty())        {            HeapNode x=Q.top(); Q.pop();            int u=x.u;            if(done[u]) continue;            done[u]=true;            for(int i=head[u];i!=-1;i=next[i])if(i != edge_id)            {                Edge &e=edges[i];                if(d[e.to] > d[u]+e.dist)                {                    d[e.to] = d[u]+e.dist;                    p[e.to]=i;//标记最短路的最后一条边                    Q.push(HeapNode(d[e.to],e.to));                }            }        }        return d[n-1];    }}DJ;int main(){    int T; scanf("%d",&T);    while(T--)    {        int n,m;        scanf("%d%d",&n,&m);        DJ.init(n);        while(m--)        {            int u,v,d;            scanf("%d%d%d",&u,&v,&d);            u--,v--;            DJ.AddEdge(u,v,d);            DJ.AddEdge(v,u,d);        }        if(DJ.dijkstra(-1)==INF) { printf("-1\n"); continue; }        vector<int> E;//保存最短路上边的序号        int ed=n-1;        while(ed!=0)        {            int edge_id = DJ.p[ed];            E.push_back(edge_id);            ed= DJ.edges[edge_id].from;        }        int ans = -1;        for(int i=0;i<E.size();i++)            ans = max(ans,DJ.dijkstra(E[i]));        if(ans==INF) printf("-1\n");        else printf("%d\n",ans);    }    return 0;}

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