HDU 4888 Redraw Beautiful Drawings 网络流 建图

题意：

给定n, m, k

以下n个整数 a

以下m个整数 b

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
, 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
;
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
, top, cur
;
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

*/