HDU 3338 Kakuro Extension 网络流

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题意：给一个N∗N的矩阵，白色的格子要求填数（范围1~9）,要求：如果有出现如028\XXX（称为上边的情况）,则表示，我这个格子下面所有的白格子中的数字之和要等于这个。如果有出现XXX\028（称为左边的情况）,则表示，我这个格子右边所有的白格子中的数字之和要等于这个。

思路：神奇的网络流。白格子需要拆为入点和出点，来控制经过白格子的流量。源点与每个左边的情况连一条流量为左边情况数字的边，每个左边的情况与他右边的白格子的入点连一条流量为INF边。每个上边的点与汇点连一条流量为上边数字的边，他下边的所有白格子出点与上边的情况，连一条流量为INF的边。每个白格子的入点和出点连一条流量为8的边。（因为流量可以为0，所以所有的流量都减去1，用0~8来表示1~9，注意的是，上边情况和左边情况的数值都要减去它所连的边的数量，最后输出结果时统一+1）

坑点：细节很多，调试起来有点烦。

http://acm.hdu.edu.cn/showproblem.php?pid=3338

/*********************************************    Problem : HDU 3338    Author  : NMfloat    InkTime (c) NM . All Rights Reserved .********************************************/#include <map>#include <set>#include <queue>#include <cmath>#include <ctime>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define rep(i,a,b)  for(int i = a ; i <= b ; i ++)#define rrep(i,a,b) for(int i = b ; i >= a ; i --)#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)#define cls(a,x)   memset(a,x,sizeof(a))#define eps 1e-8using namespace std;const int MOD = 1e9+7;typedef long long LL;int T,n,m,k;const int MAXN = 100010;  const int MAXE = 400010;  const int INF = 0x3f3f3f3f;  struct Edge{      int to,next,cap,flow;  }edge[MAXE];  int tol;  int head[MAXN];  int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];  void init(){      tol = 0;      memset(head,-1,sizeof(head));  }  void addedge(int u,int v,int w,int rw = 0){      //printf("%d->%d\t",u,v);    edge[tol].to = v;edge[tol].cap = w;edge[tol].next = head[u];      edge[tol].flow = 0;head[u] = tol ++;      edge[tol].to = u;edge[tol].cap = rw;edge[tol].next = head[v];      edge[tol].flow = 0;head[v] = tol ++;  }  int sap(int start,int end,int N){      memset(gap,0,sizeof(gap));      memset(dep,0,sizeof(dep));      memcpy(cur,head,sizeof(head));      int u = start;      pre[u] = -1;      gap[0] = N;      int ans = 0;      while(dep[start] < N){          if(u == end){              int Min = INF;              for(int i= pre[u];i!=-1;i=pre[edge[i^1].to]){                  if(Min > edge[i].cap - edge[i].flow)                      Min = edge[i].cap - edge[i].flow;              }              for(int i= pre[u];i!=-1;i=pre[edge[i^1].to]){                  edge[i].flow += Min;                  edge[i^1].flow -= Min;              }              u = start;              ans += Min;              continue;          }          bool flag = false;          int v;          for(int i=cur[u];i != -1;i = edge[i].next){              v = edge[i].to;              if(edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u]){                  flag = true;                  cur[u] = pre[v] = i;                  break;              }          }          if(flag){              u = v;              continue;          }          int Min = N;          for(int i = head[u];i != -1;i=edge[i].next){              if(edge[i].cap-edge[i].flow && dep[edge[i].to] < Min){                  Min = dep[edge[i].to];                  cur[u] = i;              }          }          gap[dep[u]]--;          if(!gap[dep[u]]) return ans;          dep[u] = Min + 1;          gap[dep[u]] ++;          if(u != start) u = edge[pre[u]^1].to;      }      return ans;  }  int Map[105][105][2];//0代表左，1代表上。int Mat[105][105][2];//0是入点，1是出点。左0，上1。int Ans[105][105];int idx ;void outMap() {    rep(i,1,n) { rep(j,1,m) printf("%d %d\t",Map[i][j][0],Map[i][j][1]); puts("");}}void outMat() {    rep(i,1,n) { rep(j,1,m) printf("%d %d\t",Mat[i][j][0],Mat[i][j][1]); puts("");}}int get_num(char s1,char s2,char s3,int flag,int x,int y) {    flag = 1 - flag;    if(s1 == 'X') Map[x][y][flag] = -1;    else if(s1 == '.') { Map[x][y][flag] = 0;Mat[x][y][flag] = idx ++;}    else { Map[x][y][flag] = (s1-'0')*100+(s2-'0')*10+(s3-'0');Mat[x][y][flag] = idx++;}}void input() {    char tmp_c[10];    idx = 1;    cls(Mat,0);    rep(i,1,n) rep(j,1,m) {        scanf("%s",tmp_c);        get_num(tmp_c[0],tmp_c[1],tmp_c[2],0,i,j);        get_num(tmp_c[4],tmp_c[5],tmp_c[6],1,i,j);    }}bool dvis[MAXN];void dfs(int u) {    int v;    for(int i = head[u] ; i != -1 ; i = edge[i].next) {        v = edge[i].to;        if(!dvis[v]) {            dvis[v] = 1;            printf("%d->%d\t",u,v);            dfs(v);        }    }}void solve() {    //建边    //outMap();puts("");    //outMat();puts("");    int start = 0 , end = idx ,totnum = idx + 1;    int tmp;    rep(i,1,n) rep(j,1,m) {        if(Map[i][j][0] > 0) { //左            tmp = 0;            rep(k,j+1,m) {                if(Map[i][k][0] == 0) {                    tmp ++;                    //printf("%d %d\n",Mat[i][j][0],Mat[i][k][0]);                    addedge(Mat[i][j][0],Mat[i][k][0],INF);                }                else break;            }            addedge(start,Mat[i][j][0],Map[i][j][0]-tmp);        }        if(Map[i][j][1] > 0) { //上            tmp = 0;            rep(k,i+1,n) {                if(Map[k][j][1] == 0) {                    tmp ++;                    addedge(Mat[k][j][1],Mat[i][j][1],INF);                }                else break;            }             addedge(Mat[i][j][1],end,Map[i][j][1]-tmp);        }        if(Map[i][j][0] == 0) { //...            addedge(Mat[i][j][0],Mat[i][j][1],8);        }    }    sap(start,end,totnum);    // cls(dvis,0);    // dvis[0] = 1;    // dfs(0);    cls(Ans,-1);    rep(i,1,n) rep(j,1,m) {        if(Map[i][j][0] == 0) {            //printf("%d->%d ",Mat[i][j][0],edge[head[Mat[i][j][0]]].to);            Ans[i][j] = edge[head[Mat[i][j][0]]].flow;        }    }    rep(i,1,n) rep(j,1,m) {            if(Ans[i][j] == -1) printf("_");            else printf("%d",Ans[i][j]+1);            if(j == m) puts("");            else printf(" ");        }}int main(void) {    //freopen("a.in","r",stdin);    //scanf("%d",&T);    //while(T--) {    while(~scanf("%d %d",&n,&m)) {        init();        input();        solve();    }    return 0;}

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