BZOJ3171: [Tjoi2013]循环格

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易知题目要求最终达到的状态是原图有若干个环，那么所有点的出度和入度都为1，费用流…

建图的话，拆点，源向左侧点连容量为1，费用0的边，代表出度，右侧点向汇连容量为1，费用0的边，代表入度，然后每个点的向四个方向的点连容量1，如果不是本来的方向费用1，否则费用0

code：

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long longusing namespace std;void up(int &x,int y){if(y>x)x=y;}void down(int &x,int y){if(y<x)x=y;}const int maxn = 1100;const int maxm = 110000;const char di[4]={'U','R','D','L'};struct edge{    int y,c,d,nex;    edge(){}    edge(int _y,int _c,int _d,int _nex){y=_y;c=_c;d=_d;nex=_nex;}}a[maxm]; int len,fir[maxn];int dir[maxn];int n,m,st,ed;void ins(int x,int y,int c,int d){    a[++len]=edge(y,c,d,fir[x]); fir[x]=len;    a[++len]=edge(x,0,-d,fir[y]); fir[y]=len;}struct node{    int x,i;    node(){}    node(int _x,int _i){x=_x;i=_i;}};bool operator <(node x,node y){return x.x>y.x;}queue<node>Q;int pos[maxn],pre[maxn],d[maxn];bool bfs(){    for(int i=1;i<=ed;i++) d[i]=INT_MAX;    d[st]=0; Q.push(node(0,st));    while(!Q.empty())    {        node x=Q.front(); Q.pop();        if(x.x>d[x.i]) continue;        for(int k=fir[x.i];k;k=a[k].nex)        {            int y=a[k].y;            if(a[k].c&&d[y]>d[x.i]+a[k].d)            {                d[y]=d[x.i]+a[k].d;                pos[y]=x.i; pre[y]=k;                Q.push(node(d[y],y));            }        }    }    return d[ed]!=INT_MAX;}int g(){    int k=INT_MAX;    for(int x=ed;x!=st;x=pos[x]) down(k,a[pre[x]].c);    for(int x=ed;x!=st;x=pos[x])    {        a[pre[x]].c-=k; a[pre[x]^1].c+=k;    }    return k*d[ed];}int main(){    char s[1000];    memset(fir,0,sizeof fir); len=1;    scanf("%d%d",&n,&m);int nm=n*m; gets(s);    int id=0;    for(int i=1;i<=n;i++)    {        for(int j=1;j<=m;j++)        {            id++;            char str; scanf("%c",&str);            for(int k=0;k<4;k++) if(str==di[k]) dir[id]=k;        }        gets(s);    }    st=id*2+1; ed=st+1;    for(int i=1;i<=id;i++) ins(st,i,1,0),ins(i+id,ed,1,0);    id=0;    for(int i=1;i<=n;i++)    {        for(int j=1;j<=m;j++)        {            id++; int idy;            //1            idy=id-m;            if(i==1) idy+=nm;            ins(id,idy+nm,1,dir[id]!=0);            //2            idy=id+1;            if(j==m) idy-=m;            ins(id,idy+nm,1,dir[id]!=1);            //3            idy=id+m;            if(i==n) idy-=nm;            ins(id,idy+nm,1,dir[id]!=2);            //4            idy=id-1;            if(j==1) idy+=m;            ins(id,idy+nm,1,dir[id]!=3);        }    }    int ret=0;    while(bfs())         ret+=g();    printf("%d\n",ret);    return 0;}

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