题意：

给定n, m, k

以下n个整数 a[n]

以下m个整数 b[n]

用数字[0,k]构造一个n*m的矩阵

若有唯一解则输出这个矩阵。若有多解输出Not Unique，若无解输出Impossible

思路：网络流，，。

n行当成n个点，m列当成m个点

从行-列连一条流量为k的边，然后源点-行连一条a[i]的边， 列-汇点 流量为b[i]

瞎了，该退役了 T^T

#include<stdio.h>

#include<string.h>

#include<iostream>

#include<math.h>

#include<algorithm>

#include<queue>

#include<vector>

using namespace std;

#define ll int

#define N 1005

#define M 200000

#define inf 107374182

#define inf64 1152921504606846976

struct Edge{

    ll from, to, cap, nex;

}edge[M*2];//注意这个一定要够大 不然会re 还有反向弧

ll head[N], edgenum;

void add(ll u, ll v, ll cap, ll rw = 0){ //假设是有向边则：add(u,v,cap); 假设是无向边则：add(u,v,cap,cap);

    Edge E = { u, v, cap, head[u]};

    edge[ edgenum ] = E;

    head[u] = edgenum ++;

    Edge E2= { v, u, rw,  head[v]};

    edge[ edgenum ] = E2;

    head[v] = edgenum ++;

}

ll sign[N];

bool BFS(ll from, ll to){

    memset(sign, -1, sizeof(sign));

    sign[from] = 0;

    queue<ll>q;

    q.push(from);

    while( !q.empty() ){

        ll u = q.front(); q.pop();

        for(ll i = head[u]; i!=-1; i = edge[i].nex)

        {

            ll v = edge[i].to;

            if(sign[v]==-1 && edge[i].cap)

            {

                sign[v] = sign[u] + 1, q.push(v);

                if(sign[to] != -1)return true;

            }

        }

    }

    return false;

}

ll Stack[N], top, cur[N];

ll Dinic(ll from, ll to){

    ll ans = 0;

    while( BFS(from, to) )

    {

        memcpy(cur, head, sizeof(head));

        ll u = from;      top = 0;

        while(1)

        {

            if(u == to)

            {

                ll flow = inf, loc;//loc 表示 Stack 中 cap 最小的边

                for(ll i = 0; i < top; i++)

                    if(flow > edge[ Stack[i] ].cap)

                    {

                        flow = edge[Stack[i]].cap;

                        loc = i;

                    }

                    for(ll i = 0; i < top; i++)

                    {

                        edge[ Stack[i] ].cap -= flow;

                        edge[Stack[i]^1].cap += flow;

                    }

                    ans += flow;

                    top = loc;

                    u = edge[Stack[top]].from;

            }

            for(ll i = cur[u]; i!=-1; cur[u] = i = edge[i].nex)//cur[u] 表示u所在能增广的边的下标

                if(edge[i].cap && (sign[u] + 1 == sign[ edge[i].to ]))break;

            if(cur[u] != -1)

            {

                Stack[top++] = cur[u];

                u = edge[ cur[u] ].to;

            }

            else

            {

                if( top == 0 )break;

                sign[u] = -1;

                u = edge[ Stack[--top] ].from;

            }

        }

    }

    return ans;

}

void init(){memset(head,-1,sizeof head);edgenum = 0;}

int n, m, k;

int a[500], b[500], suma, sumb;

int mp[505][505], dou[505][505]; //dou[0][i] i这列存在一个可增的点

int hehe(){

    if(suma != sumb)return -1;

    init();

    int from = 0, to = n+m + 10;

    for(int i = 1; i <= n; i++)

        for(int j = 1; j <= m; j++)

            add(i, n+j, k);

    for(int i = 1; i <= n; i++)

        add(from, i, a[i]);

    for(int i = 1; i <= m; i++)

        add(n+i, to, b[i]);

    int flow = Dinic(from, to);

    if(flow != suma) return -1;

    int tt = 1;

    for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++, tt+=2)

        mp[i][j] = edge[tt].cap;

    memset(dou, 0, sizeof dou);

    for(int i = 1; i <= n; i++)

    {

        for(int j = 1; j <= m; j++)

            for(int z = j+1; z <= m; z++)

            {

                bool v1=0,v2=0;

                if(mp[i][j]!=k&&mp[i][z]!=0)

                {

                    if(dou[z][j])return 0;

                    v1=1;

                }

                if(mp[i][j]!=0&&mp[i][z]!=k)

                {

                    if(dou[j][z])return 0;

                    v2=1;

                }

                if(v1)dou[j][z]=1;

                if(v2)dou[z][j]=1;

            }

    }

    return 1;

}

void input(){

    suma = sumb = 0;

    for(int i = 1; i <= n; i++)scanf("%d",&a[i]), suma += a[i];

    for(int i = 1; i <= m; i++)scanf("%d",&b[i]), sumb += b[i];

}

int main(){

    int u, v, i, j;

    while(~scanf("%d %d %d",&n,&m,&k)) {

        input();

        int ans = hehe();

        if(ans == -1)puts("Impossible");

        else if(ans == 0)puts("Not Unique");

        else

        {

            puts("Unique");

            for(i = 1; i <= n; i++)

                for(j = 1; j <= m; j++)

                    printf("%d%c",mp[i][j],j==m?'\n':' ');

        }

    }

    return 0;

}

/*

2 3 8

13 16

3 11 15

2 4 8

15 16

3 11 15 2

3 4 8

15 16 10

4 12 18 6

3 4 8

15 16 10

4 13 18 6

3 5 8

16 16 11

4 13 18 6 2

3 4 1

1 3 4

3 2 1 2

*/