NEFU 688 Word maker! 网络流

Word maker!

Time Limit 5000ms

Memory Limit 65536K

description

Probably, you have seen polyhedrons which have an English letter on each face of them. Little Ali has some of these polyhedrons and some words in his mind. We want to know how many of these words can be made by juxtaposing these polyhedrons. For instance, suppose that little Ali has one tetrahedron that letters B,Z,G,K are printed on its faces and one cube  which letters  O,O,O,O,O,O are printed on its faces. With this two polyhedrons little Ali can make “OK” and “GO”.
							

input

First line of Input contains number of the tests.
For each test case, first you are given an integer n, the number of polyhedrons and m, thenumber of words that  little Ali has. You can assume that 1<=n,m<=300. Then there are n string such that string i shows the letters which are printed on the faces of polyhedron i . After that you are given m words.
							

output

For each test case, print number of words that little Ali can make by juxtaposing his polyhedral.
							

sample_input

1
4  3
AAAA BBBB CCCCCC   DD
BAD DAD  DAB 
							

sample_output

2
							

----------------------

我承认我的图建麻烦了orz

源点连word[i]，word[i]连poly,poly连汇点。

---------------------

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>

using namespace std;

const int OO=1e9;//无穷大
const int maxm=171111;//边的最大数量，为原图的两倍
const int maxn=999;//点的最大数量

int node,src,dest,edge;//node节点数，src源点，dest汇点，edge边数
int head[maxn],work[maxn],dis[maxn],q[maxn];//head链表头，work临时表头，dis计算距离

struct edgenode{
    int to;//边的指向
    int flow;//边的容量
    int next;//链表的下一条边
}edges[maxm];

//初始化链表及图的信息
void prepare(int _node,int _src,int _dest)
{
    node=_node;
    src=_src;
    dest=_dest;
    for (int i=0;i<node;i++) head[i]=-1;
    edge=0;
}

//添加一条从u到v容量为c的边
void addedge(int u,int v,int c)
{
    edges[edge].flow=c;edges[edge].to=v;edges[edge].next=head[u];head[u]=edge++;
    edges[edge].flow=0;edges[edge].to=u;edges[edge].next=head[v];head[v]=edge++;
}

//广搜计算出每个点与源点的最短距离，如果不能到达汇点说明算法结束
bool Dinic_bfs()
{
    int u,v,r=0;
    for (int i=0;i<node;i++) dis[i]=-1;
    q[r++]=src;
    dis[src]=0;
    for (int l=0;l<r;l++)
    {
        u=q[l];
        for (int i=head[u];i!=-1;i=edges[i].next)
        {
            v=edges[i].to;
            if (edges[i].flow&&dis[v]<0)
            {//这条边必须要有剩余流量
                q[r++]=v;
                dis[v]=dis[u]+1;
                if (v==dest) return true;
            }
        }
    }
    return false;
}

//寻找可行流的增广路算法，按节点的距离来找，加快速度
int Dinic_dfs(int u,int exp)
{
    int v,tmp;
    if (u==dest) return exp;
    //work是临时链表头，这里用 i引用它，这样寻找过的边不再寻找
    for (int &i=work[u];i!=-1;i=edges[i].next)
    {
        v=edges[i].to;
        if (edges[i].flow&&dis[v]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,edges[i].flow)))>0)
        {
            edges[i].flow-=tmp;
            edges[i^1].flow+=tmp;
            //正反向边容量改变
            return tmp;
        }
    }
    return 0;
}

//求最大流直到没有可行流
int Dinic_flow()
{
    int ret=0,tmp;
    while (Dinic_bfs())
    {
        for (int i=0;i<node;i++) work[i]=head[i];
        while ( tmp=Dinic_dfs(src,OO) ) ret+=tmp;
    }
    return ret;
}

vector<string>poly;
vector<string>words;

int ans;

int main()
{
    int T;
    int n,m;
    char templar[1111];
    scanf("%d",&T);
    while (T--)
    {
        poly.clear();
        words.clear();
        ans=0;
        scanf("%d%d",&n,&m);
        for (int i=1;i<=n;i++)
        {
            scanf("%s",templar);
            poly.push_back(templar);
        }
        for (int i=1;i<=m;i++)
        {
            scanf("%s",templar);
            words.push_back(templar);
        }
        for (int i=0;i<words.size();i++)
        {
            prepare(words[i].length()+n+2,0,words[i].length()+n+1);
            for (int j=0;j<words[i].length();j++)
            {
                addedge(src,j+1,1);
                for (int k=0;k<poly.size();k++)
                {
                    if (poly[k].find(words[i][j])!=string::npos)
                    {
                        addedge(j+1,k+1+words[i].length(),1);
                    }
                }
            }
            for (int k=0;k<poly.size();k++)
            {
                addedge(k+1+words[i].length(),dest,1);
            }
            if (Dinic_flow()==words[i].length()) ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}