hdu 4975 A simple Gaussian elimination problem.（网络流，推断矩阵是否存在）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4975

3
1 1
5
5
2 2
0 10
0 10
2 2
2 2
2 2

Case #1: So simple!
Case #2: So naive!
Case #3: So young!

题意：

给出每行每列的和，问是否存在这种表格；每一个小格放的数字仅仅能是0--9。

官方题解：http://blog.sina.com.cn/s/blog_6bddecdc0102v01l.html

代码例如以下：（套用别人HDU4888的模板）

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
#define ll __int64
#define eps 1e-8
const ll Mod=(1e9+7);
const int maxn = 510;
const int maxm = 50100;

int n,m,k;
int r[maxn],c[maxn];
int ma[maxn][maxn];


const int maxnode = 10000 + 5;
const int maxedge = 2*1000000 + 5;
const int oo = 1000000000;
int node, src, dest, nedge;
int head[maxnode], point[maxedge], next1[maxedge], flow[maxedge], capa[maxedge];//point[x]==y表示第x条边连接y，head，next为邻接表，flow[x]表示x边的动态值，capa[x]表示x边的初始值
int dist[maxnode], Q[maxnode], work[maxnode];//dist[i]表示i点的等级
void init(int _node, int _src, int _dest) //初始化，node表示点的个数，src表示起点，dest表示终点
{
    node = _node;
    src = _src;
    dest = _dest;
    for (int i = 0; i < node; i++) head[i] = -1;
    nedge = 0;
}
void addedge(int u, int v, int c1, int c2) //添加一条u到v流量为c1，v到u流量为c2的两条边
{
    point[nedge] = v, capa[nedge] = c1, flow[nedge] = 0, next1[nedge] = head[u], head[u] = (nedge++);
    point[nedge] = u, capa[nedge] = c2, flow[nedge] = 0, next1[nedge] = head[v], head[v] = (nedge++);
}
bool dinic_bfs()
{
    memset(dist, 255, sizeof (dist));
    dist[src] = 0;
    int sizeQ = 0;
    Q[sizeQ++] = src;
    for (int cl = 0; cl < sizeQ; cl++)
        for (int k = Q[cl], i = head[k]; i >= 0; i = next1[i])
            if (flow[i] < capa[i] && dist[point[i]] < 0)
            {
                dist[point[i]] = dist[k] + 1;
                Q[sizeQ++] = point[i];
            }
    return dist[dest] >= 0;
}
int dinic_dfs(int x, int exp)
{
    if (x == dest) return exp;
    for (int &i = work[x]; i >= 0; i = next1[i])
    {
        int v = point[i], tmp;
        if (flow[i] < capa[i] && dist[v] == dist[x] + 1 && (tmp = dinic_dfs(v, min(exp, capa[i] - flow[i]))) > 0)
        {
            flow[i] += tmp;
            flow[i^1] -= tmp;
            return tmp;
        }
    }
    return 0;
}
int dinic_flow()
{
    int result = 0;
    while (dinic_bfs())
    {
        for (int i = 0; i < node; i++) work[i] = head[i];
        while (1)
        {
            int delta = dinic_dfs(src, oo);
            if (delta == 0) break;
            result += delta;
        }
    }
    return result;
}
//建图前,执行一遍init()；
//加边时，执行addedge(a,b,c,0),表示点a到b流量为c的边建成（注意点序号要从0開始）
//求解最大流执行dinic_flow(),返回值即为答案


bool judge(int sumrow)
{
    int flow = 1,cost = 0;
    for(int i = 1; i <= n; i++)
        for(int j = n+1; j <= n+m; j ++)
            addedge(i,j,k,0);
    flow=dinic_flow();
    if(flow != sumrow)
        return false;
    return true;
}
int main()
{
    //k为能填原图能填的数字的最大值
    int t;
    int cas = 0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        k = 9;//最多能填9
        init(n+m+2,0,n+m+1);
        int flag = 0;
        int sumrow = 0,colrow = 0;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&r[i]);
            addedge(0,i,r[i],0);
            sumrow += r[i];
            if(r[i]<0 || r[i]>m*k)
                flag = 1;
        }
        for(int j = 1; j <= m; j ++)
        {
            scanf("%d",&c[j]);
            addedge(j+n,n+m+1,c[j],0);
            colrow += c[j];
            if(c[j]<0 || c[j]>n*k)
                flag = 1;
        }
        if(sumrow != colrow)
        {
            printf("Case #%d: So naive!\n",++cas);
            continue;
        }
        if(!judge(sumrow))
            flag = 1;
        if(flag == 1)
        {
            printf("Case #%d: So naive!\n",++cas);
            continue;
        }
        memset(ma,-1,sizeof(ma));
        int i,j;
        for(i=1; i<=n; i++)
            if(r[i]==0)
                for(j=1; j<=m; j++)
                    ma[i][j]=0;
        for(j=1; j<=m; j++)
            if(c[j]==0)
                for(i=1; i<=n; i++)
                    ma[i][j]=0;
        int tt=2;
        int sum,num,temp;
        while(tt--)
        {
            for(i=1; i<=n; i++)
            {
                if(r[i]==0)
                {
                    for(j=1; j<=m; j++)
                        if(ma[i][j]==-1)
                            ma[i][j]=0;
                    continue;
                }
                sum=0;
                num=0;
                for(j=1; j<=m; j++)
                {
                    if(ma[i][j]==-1)
                    {
                        num++;
                        temp=j;
                        sum+=min(k,c[j]);
                    }
                }
                if(num==1)
                {
                    ma[i][temp]=r[i];
                    r[i]-=ma[i][temp];
                    c[temp]-=ma[i][temp];
                    continue;
                }
                else if(sum==r[i])
                {
                    for(j=1; j<=m; j++)
                    {
                        if(ma[i][j]==-1)
                        {
                            ma[i][j]=min(k,c[j]);
                            r[i]-=ma[i][j];
                            c[j]-=ma[i][j];
                        }
                    }
                }
            }
            for(j=1; j<=m; j++)
            {
                if(c[j]==0)
                {
                    for(i=1; i<=n; i++)
                        if(ma[i][j]==-1)
                            ma[i][j]=0;
                    continue;
                }
                sum=0;
                num=0;
                for(i=1; i<=n; i++)
                {
                    if(ma[i][j]==-1)
                    {
                        num++;
                        temp=i;
                        sum+=min(k,r[i]);
                    }
                }
                if(num==1)
                {
                    ma[temp][j]=c[j];
                    r[temp]-=ma[temp][j];
                    c[j]-=ma[temp][j];
                    continue;
                }
                else if(sum==c[j])
                {
                    for(i=1; i<=n; i++)
                    {
                        if(ma[i][j]==-1)
                        {
                            ma[i][j]=min(k,r[i]);
                            r[i]-=ma[i][j];
                            c[j]-=ma[i][j];
                        }
                    }
                }
            }
        }
        flag=0;
        for(i=1; i<=n; i++)
            if(r[i]!=0)
            {
                flag=1;
                break;
            }
        for(j=1; j<=m; j++)
            if(c[j]!=0)
            {
                flag=1;
                break;
            }
        if(flag==1)
            printf("Case #%d: So young!\n",++cas);
        else
        {
            printf("Case #%d: So simple!\n",++cas);
            /* for(i=1; i<=n; i++)
             {
                 for(j=1; j<m; j++)
                     printf("%d ",ma[i][j]);
                 printf("%d\n",ma[i][m]);
             }*/
        }
    }
    return 0;
}