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

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 #include<algorithm>
  5 #include<queue>
  6 using namespace std;
  7
  8 const int maxn = 10005;
  9 const int INF = (1 << 30) - 1;
 10
 11 struct Edge{
 12     int v,next,c;
 13 }e[maxn];
 14
 15 int st,ed,lev[maxn],first[maxn],now[maxn],ecnt;
 16 int r,c;
 17
 18 void init(){
 19     memset(first,-1,sizeof(first));
 20     ecnt = 0;
 21 }
 22
 23 void addedges(int u,int v,int c){
 24     e[ecnt].next = first[u];
 25     e[ecnt].v = v;
 26     e[ecnt].c = c;
 27     first[u] = ecnt++;
 28
 29     e[ecnt].next = first[v];
 30     e[ecnt].v = u;
 31     e[ecnt].c = 0;
 32     first[v] = ecnt++;
 33 }
 34
 35 bool bfs(){
 36     queue<int> q;
 37     while(!q.empty()) q.pop();
 38     q.push(st);
 39     memset(lev,-1,sizeof(lev));
 40     lev[st] = 0;
 41     while(!q.empty()){
 42         int x = q.front();q.pop();
 43         for(int i = first[x];~i;i = e[i].next){
 44             int v = e[i].v;
 45             if(lev[v] < 0 && e[i].c > 0){
 46                 lev[v] = lev[x] + 1;
 47                 q.push(v);
 48             }
 49         }
 50     }
 51     return lev[ed] != -1;
 52 }
 53
 54 int dfs(int p,int minf){
 55     if(p == ed || minf == 0) return minf;
 56     for(int &i = now[p];~i;i = e[i].next){
 57         int v = e[i].v;
 58         if(lev[v] == lev[p] + 1 && e[i].c > 0){
 59             int d = dfs(v,min(e[i].c,minf));
 60             if(d > 0){
 61                 e[i].c -= d;
 62                 e[i^1].c += d;
 63                 return d;
 64             }
 65         }
 66     }
 67     return 0;
 68 }
 69
 70 int dinic(){
 71     int max_flow = 0,p1;
 72     while(bfs()){
 73         memcpy(now,first,sizeof(first));
 74         while((p1 = dfs(st,INF)) > 0)
 75         max_flow += p1;
 76     }
 77     return max_flow;
 78 }
 79
 80 int ans[55][55];
 81
 82 int main(){
 83     int T;
 84     scanf("%d",&T);
 85     int kase = 0;
 86     while(T--){
 87         scanf("%d %d",&r,&c);
 88         init();
 89         st = 0;ed = r+c+1;
 90
 91         int last = 0,cur;
 92         for(int i = 1;i <= r;i++){
 93             scanf("%d",&cur);
 94             addedges(st,i,cur-last-c);
 95             last = cur;
 96         }
 97
 98         last = 0;
 99         for(int i = 1;i <= c;i++){
100             scanf("%d",&cur);
101             addedges(i + r,ed,cur-last-r);
102             last = cur;
103         }
104
105         for(int i = 1;i <= r;i++)
106             for(int j = 1;j <= c;j++) addedges(i,j+r,19);
107
108         int res = dinic();
109
110         for(int u = 1;u <= r;u++){
111             for(int i = first[u];~i;i = e[i].next){
112                 int v = e[i].v;
113                 ans[u][v-r] = 19 - e[i].c;
114             }
115         }
116
117         printf("Matrix %d\n",++kase);
118         for(int i = 1;i <= r;i++){
119             printf("%d",ans[i][1] + 1);
120             for(int j = 2;j <= c;j++)
121             printf(" %d",ans[i][j] + 1);
122             printf("\n");
123         }
124         if(T) puts("");
125     }
126     return 0;
127 }