[acm/icpc2016ChinaFinal][CodeforcesGym101194] Mr.Panda and TubeMaster费用流

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via. quailty
黑白染色定向
然后相当于每个点找一个后继（就能串成很多个环
每个点拆成左右两个
如果 i 是非必要点，那么左边 i 到右边 i 连一条权值 0 的边，表示我这个点连个自环（相当于就废掉了
其他的边该怎么连就怎么连
跑个最大权匹配，写成费用流就好了

#include <bits/stdc++.h>#define INF (1<<29)#define N 10000#define M 200050using namespace std;int head[N],cnt=1,tot,S,T,p[N];int dis[N],L[35][35],R[35][35],c[35][35],r[35][35],mp[35][35],inq[N];int n,m,k;int ans,fl;inline int rd() {    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}struct Edge{ int a,b,v,cost,next; }e[M],id;inline void add(int a,int b,int v,int cost) {//  printf("%d %d %d %d\n",a,b,v,cost);    e[++cnt] = (Edge){ a,b,v,cost,head[a] }, head[a] = cnt;    e[++cnt] = (Edge){ b,a,0,-cost,head[b] },head[b] = cnt;}#define cp e[i].v  #define B e[i].b  bool SPFA() {      bool flag = false;    for (int i=1;i<=tot;i++) p[i] = 0, dis[i] = -(1<<29);    dis[S] = 0; inq[S] = 1;    queue<int> q; q.push(S);      while (!q.empty()) {          int u = q.front(); q.pop(); inq[u] = 0;        if (u == T) flag = true;        for (int i=head[u];i;i=e[i].next)              if (cp > 0 && dis[u] + e[i].cost > dis[B]) {                  dis[B] = dis[u] + e[i].cost;                  p[B] = i;                if (!inq[B]) inq[B]=1, q.push(B);              }       }      return flag;  }  void mcf() {      int g = p[T] , flow = INF;      while (g) {        flow = min(flow , e[g].v);          g = p[ e[g].a ];      }    g = p[T];      while (g) {          e[g  ].v -= flow;          e[g^1].v += flow;          ans += 1LL * e[g].cost * flow;          g = p[ e[g].a ];      }    fl += flow;}void solve() {    memset(mp,0,sizeof(mp));    for (int i=1;i<=tot;i++) head[i] = 0;    for (int i=1;i<=cnt;i++) e[i] = id;    tot = 0, cnt = 1;    ans = fl = 0;    n = rd(), m = rd();    for (int i=1;i<=n;i++)        for (int j=1;j<m;j++) c[i][j] = rd();    for (int i=1;i<n;i++)        for (int j=1;j<=m;j++) r[i][j] = rd();    S = ++tot, T = ++tot;    for (int i=1;i<=n;i++)        for (int j=1;j<=m;j++)            L[i][j] = ++tot, R[i][j] = ++tot;    k = rd();    while (k--) {        int x = rd(), y = rd(); mp[x][y] = 1;    }    for (int i=1;i<=n;i++)        for (int j=1;j<=m;j++)            add(S, L[i][j], 1, 0), add(R[i][j], T, 1 ,0);    for (int i=1;i<=n;i++)        for (int j=1;j<=m;j++) if (!mp[i][j])            add(L[i][j], R[i][j], 1, 0);    for (int i=1;i<=n;i++)        for (int j=1;j<=m;j++) {            if ((i+j)&1) {                if (i+1<=n) add(L[i][j], R[i+1][j], 1, r[i][j]);                if (i-1>=1) add(L[i][j], R[i-1][j], 1, r[i-1][j]);            } else {                if (j+1<=m) add(L[i][j], R[i][j+1], 1, c[i][j]);                if (j-1>=1) add(L[i][j], R[i][j-1], 1, c[i][j-1]);            }        }    while (SPFA())        mcf();    if (fl != n*m)        puts("Impossible");    else        printf("%d\n",ans);}int main() {//  freopen("bounce.in","r",stdin);//  freopen("bounce.out","w",stdout);    int T = rd();    for (int i=1;i<=T;i++) {        printf("Case #%d: ",i);        solve();    }    return 0;}

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