Hdoj 3488 tour [maximum Classic Minimum Cost stream]

Title: hdoj 3488 tour

Question: give n vertices and m edges, and then ask you to calculate that each vertex can only be in one ring (Hamilton ring), and the minimum cost of all vertices is only one time.

Analysis

Graph creation scheme:

Split each vertex into two I and II

Then s connects all I, capacity 1, cost 0

II connect all t, capacity 1, cost 0

Connect yii to all vertices on the right. The cost is the weight and the capacity is 0.

AC code:

#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <fstream>#include <numeric>#include <iomanip>#include <bitset>#include <list>#include <stdexcept>#include <functional>#include <utility>#include <ctime>using namespace std;#define PB push_back#define MP make_pair#define Del(a,b) memset(a,b,sizeof(a))typedef vector<int> VI;typedef long long LL;const LL inf = 0x3f3f3f3f;const int N = 500;struct Node{    int from,to,cap,flow,cost;};vector<int> v[N];vector<Node> e;void add_Node(int from,int to,int cap,int cost){    e.push_back((Node)    {        from,to,cap,0,cost    });    e.push_back((Node)    {        to,from,0,0,-cost    });    int len = e.size()-1;    v[to].push_back(len);    v[from].push_back(len-1);}int vis[N],dis[N];int father[N],pos[N];bool BellManford(int s,int t,int& flow,int& cost){    Del(dis,inf);    Del(vis,0);    queue<int> q;    q.push(s);    vis[s]=1;    father[s]=-1;    dis[s] = 0;    pos[s] = inf;    while(!q.empty())    {        int f = q.front();        q.pop();        vis[f] = 0;        for(int i=0; i<v[f].size(); i++)        {            Node& tmp = e[v[f][i]];            if(tmp.cap>tmp.flow && dis[tmp.to] > dis[f] + tmp.cost)            {                dis[tmp.to] = dis[f] + tmp.cost;                father[tmp.to] = v[f][i];                pos[tmp.to] = min(pos[f],tmp.cap - tmp.flow);                if(vis[tmp.to] == 0)                {                    vis[tmp.to]=1;                    q.push(tmp.to);                }            }        }    }    if(dis[t] == inf)        return false;    flow += pos[t];    cost += dis[t]*pos[t];    for(int u = t; u!=s ; u = e[father[u]].from)    {        e[father[u]].flow += pos[t];        e[father[u]^1].flow -= pos[t];    }    return true;}int Mincost(int s,int t){    int flow = 0, cost = 0;    while(BellManford(s,t,flow,cost))    {        //printf("%d\n",cost);    }    return cost;}void Clear(int x){    for(int i=0; i<=x; i++)        v[i].clear();    e.clear();}char mp[120][120];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n,m;        scanf("%d%d",&n,&m);        int s = 0 , t = 1;        for(int i=1;i<=n;i++)        {            add_Node(s,i*2,1,0);            add_Node(i*2+1,t,1,0);        }        for(int i=0;i<m;i++)        {            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            add_Node(x*2,y*2+1,1,z);        }        int ans = Mincost(s,t);        printf("%d\n",ans);        Clear(2*n+5);    }    return 0;}

Hdoj 3488 tour [maximum Classic Minimum Cost stream]