hdu3338 Kakuro Extension 最大流

If you solved problem like this, forget it.Because you need to use a completely different algorithm to solve the following one.
Kakuro puzzle is played on a grid of "black" and "white" cells. Apart from the top row and leftmost column which are entirely black, the grid has some amount of white cells which form "runs" and some amount of black cells. "Run" is a vertical or horizontal maximal one-lined block of adjacent white cells. Each row and column of the puzzle can contain more than one "run". Every white cell belongs to exactly two runs — one horizontal and one vertical run. Each horizontal "run" always has a number in the black half-cell to its immediate left, and each vertical "run" always has a number in the black half-cell immediately above it. These numbers are located in "black" cells and are called "clues".The rules of the puzzle are simple:

1.place a single digit from 1 to 9 in each "white" cell
2.for all runs, the sum of all digits in a "run" must match the clue associated with the "run"

Given the grid, your task is to find a solution for the puzzle.

题意：给出一个特殊数独，它给出了连续横向空格和连续纵向空格的和，求数独的任意解。

将空格和它所在的连续横向空格组和连续纵向空格组建边，横向和纵向的组分别向超级源点和超级汇点建边，这样就可以跑最大流求出每个点的流量，即是答案了。

 #include<stdio.h>

 #include<string.h>

 #include<vector>

 #include<queue>

 #include<algorithm>

 using namespace std;

 const int maxm=;

 const int INF=0x7fffffff;

 struct edge{

     int from,to,f;

     edge(int a,int b,int c):from(a),to(b),f(c){}

 };

 struct dinic{

     int s,t,m;

     vector<edge>e;

     vector<int>g[maxm];

     bool vis[maxm];

     int cur[maxm],d[maxm];

     void init(int n){

         for(int i=;i<=n;i++)g[i].clear();

         e.clear();

     }

     void add(int a,int b,int c){

         e.push_back(edge(a,b,c));

         e.push_back(edge(b,a,));

         m=e.size();

         g[a].push_back(m-);

         g[b].push_back(m-);

     }

     bool bfs(){

         memset(vis,,sizeof(vis));

         queue<int>q;

         q.push(s);

         vis[s]=;

         d[s]=;

         while(!q.empty()){

             int u=q.front();

             q.pop();

             for(int i=;i<g[u].size();i++){

                 edge tmp=e[g[u][i]];

                 if(!vis[tmp.to]&&tmp.f>){

                     d[tmp.to]=d[u]+;

                     vis[tmp.to]=;

                     q.push(tmp.to);

                 }

             }

         }

         return vis[t];

     }

     int dfs(int x,int a){

         if(x==t||a==)return a;

         int flow=,f;

         for(int& i=cur[x];i<g[x].size();i++){

             edge& tmp=e[g[x][i]];

             if(d[tmp.to]==d[x]+&&tmp.f>){

                 f=dfs(tmp.to,min(a,tmp.f));

                 tmp.f-=f;

                 e[g[x][i]^].f+=f;

                 flow+=f;

                 a-=f;

                 if(a==)break;

             }

         }

         if(flow==)d[x]=-;

         return flow;

     }

     void mf(int s,int t){

         this->s=s;

         this->t=t;

         int flow=;

         while(bfs()){

             memset(cur,,sizeof(cur));

             flow+=dfs(s,INF);

         }

     }

 };

 char s[][][];

 bool vis[][];

 char ans[][];

 int main(){

     int n,m;

     while(scanf("%d%d",&n,&m)!=EOF){

         int i,j,k;

         memset(vis,,sizeof(vis));

         dinic d;

         d.init(n*m*+);

         for(i=;i<=n;i++){

             for(j=;j<=m;j++){

                 scanf("%s",s[i][j]+);

                 if(s[i][j][]=='.'){

                     vis[i][j]=;

                     d.add((i-)*m+j,(i-)*m+j+n*m,);

                 }

                 else ans[i][j]='_';

             }

         }

         int t=n*m*+;

         for(i=;i<=n;i++){

             for(j=;j<=m;j++){

                 if(!vis[i][j]){

                     if(s[i][j][]!='X'){

                         int tmp=(s[i][j][]-'')*+(s[i][j][]-'')*+(s[i][j][]-''),num=(i-)*m+j;

                         for(k=i+;k<=n&&vis[k][j];k++){

                             tmp--;

                             d.add(num,(k-)*m+j,INF);

                         }

                         d.add(,num,tmp);

                     }

                     if(s[i][j][]!='X'){

                         int tmp=(s[i][j][]-'')*+(s[i][j][]-'')*+(s[i][j][]-''),num=(i-)*m+j+n*m;

                         for(k=j+;k<=m&&vis[i][k];k++){

                             tmp--;

                             d.add((i-)*m+k+n*m,num,INF);

                         }

                         d.add(num,t,tmp);

                     }

                 }

             }

         }

         d.mf(,t);

         for(i=;i<d.e.size();i++){

             if(d.e[i].from<=n*m&&d.e[i].to>n*m){

             int x=d.e[i].from/m+,y=d.e[i].from%m;

             if(y==){y=m;x--;}

                 ans[x][y]=-d.e[i].f++'';

             }

         }

         for(i=;i<=n;i++){

             for(j=;j<=m;j++){

                 printf("%c",ans[i][j]);

                 if(j==m)printf("\n");

                 else printf(" ");

             }

         }

     }

     return ;

 }

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hdu3338 Kakuro Extension 最大流

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