POJ 2455 Secret Milking Machine 最大流建模

题目链接

http://poj.org/problem?id=2455

题意

某人为了隐藏他的挤奶机，可能是为了不让别人跟踪，他决定从1~n，然后走密道回到1，再从1到n，走t次，且每条路径不同，每次尽量走路径最短的路，问他在t次走的过程中，两个路标之间的最大距离。

思路

二分枚举最大边权。如果两点的边权小于等于枚举的量，之间建立一条容量为1的边，若有重边满足条件，容量加1，最后从1点跑最大流，判断最大流是否大于等于t。边是双向的，所以反向边的残量不为0

#include<cstdio>#include<queue>#include<iostream>#include<vector>#include<map>#include<cstring>#include<string>#include<set>#include<stack>#include<algorithm>#define cle(a) memset(a,0,sizeof(a))#define inf(a) memset(a,0x3f,sizeof(a))#define ll long long#define Rep(i,a,n) for(int i=a;i<=n;i++)using namespace std;#define INF2 9223372036854775807llconst int INF = ( 2e9 ) + 2;const ll maxn =  410;const int maxm = 210;ll maxx;int n,m,t;int mp2[maxn][maxn];ll mp[maxn][maxn];struct edge{    int from,to,c;    edge(int getu,int getv,int getc)    {        from=getu;        to=getv;        c=getc;    }};vector<edge> e;struct Dinic{    int s,t,n;    int d[maxn];    int cur[maxn];    vector<edge> e;    vector<int> g[maxn];    void addedge(int u,int v,int c)    {        e.push_back(edge(u,v,c));        e.push_back(edge(v,u,c));        int m=e.size();        g[u].push_back(m-2);        g[v].push_back(m-1);    }    void init()    {        e.clear();        for(int i=0; i<=maxn; i++)            g[i].clear();    }    bool bfs()    {        queue<int> q;        q.push(s);        memset(d,0,sizeof(d));        d[s]=1;        while(!q.empty())        {            int u=q.front();            q.pop();            for(int i=0; i<g[u].size(); i++)            {                int v=e[g[u][i]].to;                int c=e[g[u][i]].c;                if(!d[v]&&c)                {                    d[v]=d[u]+1;                    q.push(v);                }            }        }        return d[t];    }    int dfs(int u,int maxf,int t)    {        if(u==t)return maxf;        int ret=0;        for(int &i=cur[u]; i<g[u].size(); i++)        {            int c=e[g[u][i]].c;            int v=e[g[u][i]].to;            int f;            if(d[u]+1==d[v]&&c)            {                f=dfs(v,min(maxf-ret,c),t);                e[g[u][i]].c-=f;                e[g[u][i]^1].c+=f;                ret+=f;                if(ret==maxf)return ret;            }        }        return ret;    }    int maxflow(int s,int t)    {        this->s=s;        this->t=t;        int flow=0;        while(bfs())        {            memset(cur,0,sizeof(cur));            int temp=dfs(s,INF,t);            flow+=temp;        }        return flow;    }} dinic;bool check(int t,int n,int T){    for(int i=0; i<e.size(); i++)    {        if(e[i].c<=t)            dinic.addedge(e[i].from,e[i].to,1);    }    int temp=dinic.maxflow(1,n);    dinic.init();    return temp>=T;}int main(){    while(~scanf("%d%d%d",&n,&m,&t))    {        memset(mp,0,sizeof(mp));        e.clear();        for(int i=0; i<m; i++)        {            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            e.push_back(edge(u,v,w));            mp[u][v]=1;            mp[v][u]=1;        }        int l=0,r=1000010;        while(l<r)        {            int mid=(l+r)>>1;            if(check(mid,n,t))                r=mid;            else                l=mid+1;        }        cout<<l<<endl;    }}