loj 1167(二分+最大流）

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=26881

思路：我们可以二分最大危险度，然后建图，由于每个休息点只能用一次，就要拆点，将每个休息点拆点，边容量为1，代表只能使用一次，然后跑最大流验证。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
using namespace std;
#define MAXN 222
#define MAXM 222222
#define inf 1<<30
#define FILL(a,b) memset(a,b,sizeof(a))

struct Edge{
    int v,cap,next;
}edge[MAXM];

int n,m,vs,vt,k,NV,NE;
int head[MAXN];

void Insert(int u,int v,int cap)
{
    edge[NE].v=v;
    edge[NE].cap=cap;
    edge[NE].next=head[u];
    head[u]=NE++;

    edge[NE].v=u;
    edge[NE].cap=0;
    edge[NE].next=head[v];
    head[v]=NE++;
}

int level[MAXN],gap[MAXN];
void bfs(int vt)
{
    FILL(level,-1);
    FILL(gap,0);
    queue<int>que;
    level[vt]=0;
    gap[0]++;
    que.push(vt);
    while(!que.empty()){
        int u=que.front();
        que.pop();
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].v;
            if(level[v]!=-1)continue;
            level[v]=level[u]+1;
            gap[level[v]]++;
            que.push(v);
        }
    }
}

int pre[MAXN],cur[MAXN];
int SAP(int vs,int vt)
{
    bfs(vt);
    memcpy(cur,head,sizeof(head));
    int u=pre[vs]=vs,aug=inf,maxflow=0;
    gap[0]=NV;
    while(level[vs]<NV){
        bool flag=false;
        for(int &i=cur[u];i!=-1;i=edge[i].next){
            int v=edge[i].v;
            if(edge[i].cap>0&&level[u]==level[v]+1){
                flag=true;
                aug=min(aug,edge[i].cap);
                pre[v]=u;
                u=v;
                if(v==vt){
                    maxflow+=aug;
                    for(u=pre[u];v!=vs;v=u,u=pre[u]){
                        edge[cur[u]].cap-=aug;
                        edge[cur[u]^1].cap+=aug;
                    }
                    aug=inf;
                }
                break;
            }
        }
        if(flag)continue;
        int minlevel=NV;
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].v;
            if(edge[i].cap>0&&level[v]<minlevel){
                minlevel=level[v];
                cur[u]=i;
            }
        }
        if(--gap[level[u]]==0)break;
        level[u]=minlevel+1;
        gap[level[u]]++;
        u=pre[u];
    }
    return maxflow;
}

struct Path{
    int u,v,d;
}path[MAXM];

void Build(int limit)
{
    NE=0;
    memset(head,-1,sizeof(head));
    vs=0,vt=2*n+3,NV=2*n+4;
    for(int i=0;i<m;i++){
        if(path[i].d<=limit)Insert(path[i].u+n+2,path[i].v,1);
    }
    for(int i=1;i<=n;i++)Insert(i,i+n+2,1);
    Insert(vs,vs+n+2,11);
    Insert(n+1,vt,11);
}


int main()
{
    int _case,u,v,d,low,high,mid,ans,t=1;
    scanf("%d",&_case);
    while(_case--){
        scanf("%d%d",&n,&m);
        low=inf,high=0;
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&path[i].u,&path[i].v,&path[i].d);
            if(path[i].u>path[i].v)swap(path[i].u,path[i].v);
            //cout<<path[i].u<<"&&&&"<<path[i].v<<endl;
            low=min(low,path[i].d);
            high=max(high,path[i].d);
        }
       // cout<<low<<"**"<<high<<endl;
        scanf("%d",&k);
        ans=inf;
        while(low<=high){
            mid=(low+high)>>1;
            Build(mid);
            if(SAP(vs,vt)>=k){
                ans=mid;
                high=mid-1;
            }else
                low=mid+1;
        }
        printf("Case %d: ",t++);
        if(ans!=inf){
            printf("%d
",ans);
        }else
            puts("no solution");
    }
    return 0;
}