poj 2455

题意:由一个点走到另一个点,中间的点可以重复到达,但边只能经过一次,问T条边不重复的路径里,最长的边的最小值.

分析:由于点是可以重用的,因此不必拆点.这道题有重边,而且重边都必须保留,因为点是可以重复走的.二分最长的边长,用最大流来验证.

#include <cstdio>
#include <cstring>
#include <algorithm>
const int MAXN=205;//点数的最大值
const int MAXM=160005;//边数的最大值
const int INF=0x3fffffff;

struct Node
{
    int from,to,next;
    int cap;
}edge[MAXM];

int tol;
int head[MAXN];
int dep[MAXN];
int gap[MAXN];//gap[x]=y :说明残留网络中dep[i]==x的个数为y
int n;//n是总的点的个数，包括源点和汇点
Node container[40005];

void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int w)
{
    edge[tol].from=u;
    edge[tol].to=v;
    edge[tol].cap=w;
    edge[tol].next=head[u];
    head[u]=tol++;
    edge[tol].from=v;
    edge[tol].to=u;
    edge[tol].cap=0;
    edge[tol].next=head[v];
    head[v]=tol++;
}
void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0]=1;
    int que[MAXN];
    int front,rear;
    front=rear=0;
    dep[end]=0;
    que[rear++]=end;
    while(front!=rear)
    {
        int u=que[front++];
        if(front==MAXN)front=0;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(dep[v]!=-1)continue;
            que[rear++]=v;
            if(rear==MAXN)rear=0;
            dep[v]=dep[u]+1;
            ++gap[dep[v]];
        }
    }
}
int SAP(int start,int end)
{
    int res=0;
    BFS(start,end);
    int cur[MAXN];
    int S[MAXN];
    int top=0;
    memcpy(cur,head,sizeof(head));
    int u=start;
    int i;
    while(dep[start]<n)
    {
        if(u==end)
        {
            int temp=INF;
            int inser;
            for(i=0;i<top;i++)
               if(temp>edge[S[i]].cap)
               {
                   temp=edge[S[i]].cap;
                   inser=i;
               }
            for(i=0;i<top;i++)
            {
                edge[S[i]].cap-=temp;
                edge[S[i]^1].cap+=temp;
            }
            res+=temp;
            top=inser;
            u=edge[S[top]].from;
        }
        if(u!=end&&gap[dep[u]-1]==0)//出现断层，无增广路
          break;
        for(i=cur[u];i!=-1;i=edge[i].next)
           if(edge[i].cap!=0&&dep[u]==dep[edge[i].to]+1)
             break;
        if(i!=-1)
        {
            cur[u]=i;
            S[top++]=i;
            u=edge[i].to;
        }
        else
        {
            int min=n;
            for(i=head[u];i!=-1;i=edge[i].next)
            {
                if(edge[i].cap==0)continue;
                if(min>dep[edge[i].to])
                {
                    min=dep[edge[i].to];
                    cur[u]=i;
                }
            }
            --gap[dep[u]];
            dep[u]=min+1;
            ++gap[dep[u]];
            if(u!=start)u=edge[S[--top]].from;
        }
    }
    return res;
}
int main()
{
    int N,P,T;
    while(~scanf("%d%d%d",&N,&P,&T))
    {
        n = N;
        int left = INF,right = 0;
        for(int i = 0;i < P;i++)
        {
            scanf("%d%d%d",&container[i].from,&container[i].to,&container[i].cap);
            left = std::min(container[i].cap,left);
            right = std::max(container[i].cap,right);
        }
        left--;
        while(left + 1 < right)
        {
            int mid = (left + right) / 2;
            init();
            for(int i = 0;i < P;i++)
            {
                if(container[i].cap <= mid)
                {
                    addedge(container[i].from,container[i].to,1);
                    addedge(container[i].to,container[i].from,1);
                }
            }
            if(SAP(1,n) >= T)
            {
                right = mid;
            }
            else
            {
                left = mid;
            }
        }
        printf("%d
",right);
    }
    return 0;
}