vector 的注意事项 （poj2195 小人进房子 最小费用最大流）

下图分别是用50w和20wsize的vector跑出来的结果，vector.clear()耗时太长！尽量开小

附poj2195ac代码

    #include <cstdio>    #include <iostream>    #include <algorithm>    #include <vector>    #include <cstring>    #include <queue>    #include <string>    #include <map>    using namespace std;    #define Del(a,b) memset(a,b,sizeof(a))    const int inf = 0x3f3f3f3f;    const int N=50000;    struct Point{        int x, y;        Point():x(0), y(0){};        Point(int x0, int y0): x(x0),y(y0){};    }fih[50000], fim[50000];    struct Node    {        int from,to,cap,flow,cost;    };    vector<Node> e;    vector<int> v[N];    int vis[N],dis[N];    int p[N],a[N];  //p保存father，a保存cap    void Clear(int x)    {        for(int i=0;i<=x;i++)            v[i].clear();        e.clear();    }    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);    }    bool BellmanFord(int s,int t,int& flow,int& cost)    {        Del(dis,inf);        Del(vis,0);        dis[s] = 0;        vis[s] = 1;        p[s] = 0;        a[s] = inf;        queue<int> q;        q.push(s);        while(!q.empty())        {            int u = q.front();            q.pop();            vis[u] = 0;            for(int i=0; i<v[u].size(); i++)            {                Node& g = e[v[u][i]];                if(g.cap>g.flow && dis[g.to] > dis[u]+g.cost)                {                    dis[g.to] = dis[u] + g.cost;                    p[g.to] = v[u][i];  //保存前驱                    a[g.to] = min(a[u],g.cap-g.flow);                    if(!vis[g.to])                    {                        q.push(g.to);                        vis[g.to]=1;                    }                }            }        }        if(dis[t] == inf)            return false;        flow += a[t];        cost += dis[t]*a[t];///可能需要修改，本题下同一条路径上的每个单位流量都要消耗        int u = t;        while(u!=s)        {            e[p[u]].flow += a[t];            e[p[u]^1].flow -= a[t];            u = e[p[u]].from;        }        //cout<<"BF Done"<<endl;        return true;    }    int Min_Cost(int s,int t)    {        int flow=0,cost = 0;        while(BellmanFord(s,t,flow,cost));        //cout<<"MC Done"<<endl;        return cost;    }    int main()    {    int n,m,k;  //n 店主 k 物品 M 供应商    while((~scanf("%d%d",&n,&m)) && (m+n))  //多组测试数据，mnk均=0时输入结束    {        int s = 0, totm=0, toth=0;        for (int i=0; i<n; i++){            string rubbish;getline(cin, rubbish);        for (int j=0; j<m; j++)        {            char c;            scanf("%c", &c);            //cout<<"c="<<c<<endl;            if (c=='H')                fih[++toth]=Point(i,j);            else if(c=='m')                fim[++totm]=Point(i,j);        }        }        //s=0, 1..totm, totm+1..totm+toth, t=toth+totm+1;        for (int i=1; i<=totm; i++)            add_Node(0, i, 1, 0);        for (int i=1; i<=totm; i++)        for (int j=1; j<=toth; j++)            add_Node(i,j+totm,1, abs(fih[i].x-fim[j].x)+abs(fih[i].y-fim[j].y));        for (int i=totm+1; i<=totm+toth; i++)            add_Node(i, totm+toth+1, 1, 0);        cout<<Min_Cost(0, totm+toth+1)<<endl;        Clear(totm+toth+1);    }    return 0;    }