uva 11082 Matrix Decompressing 【 最大流 】

只看题目的话~~怎么也看不出来是网络流的题目的说啊~~~~

建图好神奇~~

最开始不懂---后来看了一下这篇--

http://www.cnblogs.com/AOQNRMGYXLMV/p/4280727.html

建立源点 st = 0,汇点 ed = r+c

因为正整数的范围是1到20，而流量可以是0，所以先将矩阵里面的每个数减去1，到最后输出答案的时候再加上1

把每一行看做一个节点 x,编号为1到r

把每一列看做一个节点y,编号为r+1到r+c

st到x连边，容量为 Ai '- c

y到ed连边，容量为Bi' - r

x到y连边，容量为19

这样再用一下 dinic

节点xi到yj的流量就是格子(i,j) - 1之后的值

这句话最开始非常非常的不理解-------

后来搜了下题解

因为yj的流量来自于各个x

同样的，xi的流量流向各个y,所以（i，j）这个格子的流量来源是xi

大概像这个图这样

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<queue>
 using namespace std;

 const int maxn = ;
 const int INF = ( << ) - ;

 struct Edge{
     int v,next,c;
 }e[maxn];

 int st,ed,lev[maxn],first[maxn],now[maxn],ecnt;
 int r,c;

 void init(){
     memset(first,-,sizeof(first));
     ecnt = ;
 }

 void addedges(int u,int v,int c){
     e[ecnt].next = first[u];
     e[ecnt].v = v;
     e[ecnt].c = c;
     first[u] = ecnt++;

     e[ecnt].next = first[v];
     e[ecnt].v = u;
     e[ecnt].c = ;
     first[v] = ecnt++;
 }

 bool bfs(){
     queue<int> q;
     while(!q.empty()) q.pop();
     q.push(st);
     memset(lev,-,sizeof(lev));
     lev[st] = ;
     while(!q.empty()){
         int x = q.front();q.pop();
         for(int i = first[x];~i;i = e[i].next){
             int v = e[i].v;
             if(lev[v] <  && e[i].c > ){
                 lev[v] = lev[x] + ;
                 q.push(v);
             }
         }
     }
     return lev[ed] != -;
 }

 int dfs(int p,int minf){
     if(p == ed || minf == ) return minf;
     for(int &i = now[p];~i;i = e[i].next){
         int v = e[i].v;
         if(lev[v] == lev[p] +  && e[i].c > ){
             int d = dfs(v,min(e[i].c,minf));
             if(d > ){
                 e[i].c -= d;
                 e[i^].c += d;
                 return d;
             }
         }
     }
     return ;
 }

 int dinic(){
     int max_flow = ,p1;
     while(bfs()){
         memcpy(now,first,sizeof(first));
         while((p1 = dfs(st,INF)) > )
         max_flow += p1;
     }
     return max_flow;
 }

 int ans[][];

 int main(){
     int T;
     scanf("%d",&T);
     int kase = ;
     while(T--){
         scanf("%d %d",&r,&c);
         init();
         st = ;ed = r+c+;

         int last = ,cur;
         for(int i = ;i <= r;i++){
             scanf("%d",&cur);
             addedges(st,i,cur-last-c);
             last = cur;
         }

         last = ;
         for(int i = ;i <= c;i++){
             scanf("%d",&cur);
             addedges(i + r,ed,cur-last-r);
             last = cur;
         }

         for(int i = ;i <= r;i++)
             for(int j = ;j <= c;j++) addedges(i,j+r,);

         int res = dinic();

         for(int u = ;u <= r;u++){
             for(int i = first[u];~i;i = e[i].next){
                 int v = e[i].v;
                 ans[u][v-r] =  - e[i].c;
             }
         }

         printf("Matrix %d\n",++kase);
         for(int i = ;i <= r;i++){
             printf("%d",ans[i][] + );
             for(int j = ;j <= c;j++)
             printf(" %d",ans[i][j] + );
             printf("\n");
         }
         if(T) puts("");
     }
     return ;
 }