Codeforces 316C2 棋盘模型

Let’s move from initial matrix to the bipartite graph. The matrix elements (i, j) for which i + j are even should be place to one part, the matrix elements (i, j) for which i + j are uneven should be place to another part. The edges are corresponding to squares which are situated side by side. After that let’s weigh the edges. The edges which connect equal elements of matrix have weights 0, for unequal elements – weight 1. After that the problem is reduced to finding of the maximum independent edge set with minimal weight. Substantiation of above-stated is following: an answer to the problem represents a partitioning of initial matrix for pairs. Note that for any partitioning minimal number of changing matrix elements corresponds to the number of pairs on unequal elements. So in the optimal partitioning the number of pairs of equal elements is maximum. For solving minimum-cost flow problem is needed to use some effective algorithm. For example, Dijkstra algorithm with heap adding conversion of edges weights from Johnson's algorithm.

#include<iostream>

#include<algorithm>

#include<cstdio>

#include<cstring>

using namespace std;

const int maxn = 1e5+11;

const int oo = 0x3f3f3f3f;

int to[maxn<<1],cost[maxn<<1],cap[maxn<<1],flow[maxn<<1],nxt[maxn<<1];

int head[maxn],tot;

void init(){

    memset(head,-1,sizeof head);

    tot=0;

}

void add(int u,int v,int c,int w){

    to[tot]=v;

    cap[tot]=c;

    flow[tot]=0;

    cost[tot]=w;

    nxt[tot]=head[u];

    head[u]=tot++;

    swap(u,v);

    to[tot]=v;

    cap[tot]=0;

    flow[tot]=0;

    cost[tot]=-w;

    nxt[tot]=head[u];

    head[u]=tot++;

}

struct QUEUE{

    int que[maxn];

    int front,rear;

    void init(){front=rear=0;}

    void push(int x){que[rear++]=x;}

    int pop(){return que[front++];}

    bool empty(){return front==rear;}

}que;

int n,m,s,t;

bool vis[maxn];

int pre[maxn],dis[maxn];

bool spfa(){

    que.init();

    memset(vis,0,sizeof vis);

    memset(pre,-1,sizeof pre);

    memset(dis,oo,sizeof dis);

    que.push(s);vis[s]=1;dis[s]=0;

    while(!que.empty()){

        int u=que.pop(); vis[u]=0;

        for(int i = head[u]; ~i; i = nxt[i]){

            int v=to[i],c=cap[i],f=flow[i],w=cost[i];

            if(c>f&&dis[v]>dis[u]+w){

                dis[v]=dis[u]+w;

                pre[v]=i;

                if(!vis[v]){

                    que.push(v);

                    vis[v]=1;

                }

            }

        }

    }

    if(dis[t]==oo) return 0;

    else return 1;

}

int mcmf(){

    int mc=0,mf=0;

    while(spfa()){

        int tf=oo+1;

        for(int i = pre[t]; ~i; i = pre[to[i^1]]){

            tf=min(tf,cap[i]-flow[i]);

        }

        mf+=tf;

        for(int i = pre[t]; ~i; i = pre[to[i^1]]){

            flow[i]+=tf;

            flow[i^1]-=tf;

        }

        mc+=dis[t]*tf;

    }

    return mc;

}

#define rep(i,j,k) for(int i = j; i <= k; i++)

#define repp(i,j,k) for(int i = j; i < k; i++)

#define repe(i,u) for(int i = head[u]; ~i; i = nxt[i])

#define scan(a) scanf("%d",&a)

#define scann(a,b) scanf("%d%d",&a,&b)

#define scannn(a,b,c) scanf("%d%d%d",&a,&b,&c)

#define println(a) printf("%d\n",a)

#define printbk(a) printf("%d ",a)

#define print(a) printf("%d",a)

int G[100][100],R,C,n1;

inline int ID(int i,int j){

    return (i-1)*C+j;

}

inline int chai(int x){

    return R*C+x;

}

inline bool black(int i,int j){

    return i+j &1;

}

int dx[]={0,1,0,-1};

int dy[]={1,0,-1,0};

int main(){

    while(scann(R,C)!=EOF){

        init();

        rep(i,1,R) rep(j,1,C) scan(G[i][j]);

        s=R*C+1;t=s+1;n=t;

        rep(i,1,R){

            rep(j,1,C){

                if(!black(i,j))continue;

                rep(d,0,3){

                    int xx=i+dx[d],yy=j+dy[d];

                    if(xx<1||xx>R||yy<1||yy>C)continue;

                    if(G[xx][yy]==G[i][j]){

                        add(ID(i,j),ID(xx,yy),1,0);

                    }

                    else{

                        add(ID(i,j),ID(xx,yy),1,1);

                    }

                }

            }

        }

        rep(i,1,R) rep(j,1,C){

            if(black(i,j)) add(s,ID(i,j),1,0);

            else add(ID(i,j),t,1,0);

        }

        println(mcmf());

    }

    return 0;

}

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Codeforces 316C2 棋盘模型

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