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 = 10005;
const int INF = (1 << 30) - 1;

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,-1,sizeof(first));
    ecnt = 0;
}

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 = 0;
    first[v] = ecnt++;
}

bool bfs(){
    queue<int> q;
    while(!q.empty()) q.pop();
    q.push(st);
    memset(lev,-1,sizeof(lev));
    lev[st] = 0;
    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] < 0 && e[i].c > 0){
                lev[v] = lev[x] + 1;
                q.push(v);
            }
        }
    }
    return lev[ed] != -1;
}

int dfs(int p,int minf){
    if(p == ed || minf == 0) return minf;
    for(int &i = now[p];~i;i = e[i].next){
        int v = e[i].v;
        if(lev[v] == lev[p] + 1 && e[i].c > 0){
            int d = dfs(v,min(e[i].c,minf));
            if(d > 0){
                e[i].c -= d;
                e[i^1].c += d;
                return d;
            }
        }
    }
    return 0;
}

int dinic(){
    int max_flow = 0,p1;
    while(bfs()){
        memcpy(now,first,sizeof(first));
        while((p1 = dfs(st,INF)) > 0)
        max_flow += p1;
    }
    return max_flow;
}

int ans[55][55];

int main(){
    int T;
    scanf("%d",&T);
    int kase = 0;
    while(T--){
        scanf("%d %d",&r,&c);
        init();
        st = 0;ed = r+c+1;
        
        int last = 0,cur;        
        for(int i = 1;i <= r;i++){
            scanf("%d",&cur);
            addedges(st,i,cur-last-c);
            last = cur;
        }
        
        last = 0;
        for(int i = 1;i <= c;i++){
            scanf("%d",&cur);
            addedges(i + r,ed,cur-last-r);
            last = cur;
        }
        
        for(int i = 1;i <= r;i++)
            for(int j = 1;j <= c;j++) addedges(i,j+r,19);
        
        int res = dinic();
        
        for(int u = 1;u <= r;u++){
            for(int i = first[u];~i;i = e[i].next){
                int v = e[i].v;
                ans[u][v-r] = 19 - e[i].c;
            }
        }
        
        printf("Matrix %d
",++kase);
        for(int i = 1;i <= r;i++){
            printf("%d",ans[i][1] + 1);
            for(int j = 2;j <= c;j++)
            printf(" %d",ans[i][j] + 1);
            printf("
");
        }
        if(T) puts("");
    }
    return 0;
}