Redraw Beautiful Drawings
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2909 Accepted Submission(s):
942
Problem Description
Alice and Bob are playing together. Alice is crazy
about art and she has visited many museums around the world. She has a good
memory and she can remember all drawings she has seen.
about art and she has visited many museums around the world. She has a good
memory and she can remember all drawings she has seen.
Today Alice
designs a game using these drawings in her memory. First, she matches K+1 colors
appears in the picture to K+1 different integers(from 0 to K). After that, she
slices the drawing into grids and there are N rows and M columns. Each grid has
an integer on it(from 0 to K) representing the color on the corresponding
position in the original drawing. Alice wants to share the wonderful drawings
with Bob and she tells Bob the size of the drawing, the number of different
colors, and the sum of integers on each row and each column. Bob has to redraw
the drawing with Alice's information. Unfortunately, somtimes, the information
Alice offers is wrong because of Alice's poor math. And sometimes, Bob can work
out multiple different drawings using the information Alice provides. Bob gets
confused and he needs your help. You have to tell Bob if Alice's information is
right and if her information is right you should also tell Bob whether he can
get a unique drawing.
Input
The input contains mutiple testcases.
For each
testcase, the first line contains three integers N(1 ≤ N ≤ 400) , M(1 ≤ M ≤ 400)
and K(1 ≤ K ≤ 40).
N integers are given in the second line representing the
sum of N rows.
M integers are given in the third line representing the sum of
M columns.
The input is terminated by EOF.
Output
For each testcase, if there is no solution for Bob,
output "Impossible" in one line(without the quotation mark); if there is only
one solution for Bob, output "Unique" in one line(without the quotation mark)
and output an N * M matrix in the following N lines representing Bob's unique
solution; if there are many ways for Bob to redraw the drawing, output "Not
Unique" in one line(without the quotation mark).
output "Impossible" in one line(without the quotation mark); if there is only
one solution for Bob, output "Unique" in one line(without the quotation mark)
and output an N * M matrix in the following N lines representing Bob's unique
solution; if there are many ways for Bob to redraw the drawing, output "Not
Unique" in one line(without the quotation mark).
Sample Input
2 2 4
4 2
4 2
4 2 2
2 2 5 0
5 4
1 4 3
9
1 2 3 3
Sample Output
Not Unique
Impossible
Unique
1 2 3 3
题意：在N*M的表格内填入数字，使得每行每列相加等于对应的值，且填入的数字小于K值；
如果不存在这样的表格，输出Impossible；如果存在多个符合条件的表格，输出Not Unique，否则，如果表格唯一的话，输出Unique，并输出表格；
思路：网络流可以做，当然网络流+最大流更好！
网络流（687MS）
#include<iostream>//网络流
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<vector>
#include<queue>
#include<cmath>
#define maxn 1<<29
using namespace std;
struct edge
{
int from,to,cap,flow;
};
vector<int>g[];
vector<edge>edges;
int m,n,ma;
bool vis[];
int d[];
int cur[];
int fl[][];
bool cc[][];
void init()
{
edges.clear();
int mm=m+n+;
for(int i=;i<=mm;i++)g[i].clear();
}
void add(int u,int v,int c)
{
edges.push_back((edge){u,v,c,});
g[u].push_back(edges.size()-);
edges.push_back((edge){v,u,,});
g[v].push_back(edges.size()-);
}
bool bfs(int s,int t)
{
memset(vis,,sizeof(vis));
queue<int>q;
q.push(s);
d[s]=;
vis[s]=;
while(!q.empty())
{
int u=q.front();
q.pop();
int size=g[u].size();
for(int i=;i<size;i++)
{
edge &e=edges[g[u][i]];
if(!vis[e.to]&&e.cap>e.flow)
{
vis[e.to]=;
d[e.to]=d[u]+;
q.push(e.to);
}
}
}
return vis[t];
}
int dfs(int u,int t,int mi)
{
if(u==t||mi==)return mi;
int flow=,f;
int size=g[u].size();
for(int &i=cur[u];i<size;i++)
{
edge &e=edges[g[u][i]];
if(d[u]+==d[e.to]&&(f=dfs(e.to,t,min(mi,e.cap-e.flow)))>)
{
e.flow+=f;
edges[g[u][i]^].flow-=f;
flow+=f;
mi-=f;
if(mi==)break;
}
}
return flow;
}
int dinic(int s,int t)
{
int flow=;
while(bfs(s,t))
{
memset(cur,,sizeof(cur));
flow+=dfs(s,t,maxn);
}
return flow;
}
bool go()
{
for(int i=;i<=n;i++)
{
int size=g[i].size();
for(int j=;j<size;j++)
{
edge &e=edges[g[i][j]];
if(e.to>n&&e.to<=m+n)
{
//cout<<e.from<<" "<<e.to<<" "<<e.flow<<endl;
fl[i][e.to-n]=e.flow;
}
}
}
memset(cc,,sizeof(cc));
for(int i=;i<=n;i++)
{
for(int j=;j<=m;j++)
{
for(int k=j+;k<=m;k++)
{
bool v1=,v2=;
if(fl[i][j]!=ma&&fl[i][k]!=)
{
if(cc[k][j])return true;
v1=;
}
if(fl[i][j]!=&&fl[i][k]!=ma)
{
if(cc[j][k])return true;
v2=;
}
if(v1)cc[j][k]=;
if(v2)cc[k][j]=;
}
}
}
return false;
}
int main()
{
int u,v,c;
int s1,s2;
while(scanf("%d%d%d",&n,&m,&ma)!=EOF)
{
init();
s1=s2=;
for(int i=;i<=n;i++)
{
scanf("%d",&c);
add(,i,c);
s1+=c;
for(int j=;j<=m;j++)
{
add(i,n+j,ma);
}
}
for(int i=;i<=m;i++)
{
scanf("%d",&c);
add(n+i,m+n+,c);
s2+=c;
}
int ans=dinic(,m+n+);
if(ans!=s1||ans!=s2)printf("Impossible\n");
else if(go())printf("Not Unique\n");
else
{
printf("Unique\n");
for(int i=;i<=n;i++)
{
for(int j=;j<=m;j++)
{
printf("%d",fl[i][j]);
if(j==m)printf("\n");
else printf(" ");
}
}
}
}
return ;
}
网络流+最大流（234MS）
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#include <numeric>
using namespace std;
typedef long long LL;
const int MAXN = ;
const int MAXV = MAXN << ;
const int MAXE = * MAXN * MAXN;
const int INF = 0x3f3f3f3f;
struct ISAP
{
int head[MAXV], cur[MAXV], gap[MAXV], dis[MAXV], pre[MAXV];
int to[MAXE], next[MAXE], flow[MAXE];
int n, ecnt, st, ed;
void init(int n)
{
this->n = n;
memset(head + , -, n * sizeof(int));
ecnt = ;
}
void add_edge(int u, int v, int c)
{
to[ecnt] = v;
flow[ecnt] = c;
next[ecnt] = head[u];
head[u] = ecnt++;
to[ecnt] = u;
flow[ecnt] = ;
next[ecnt] = head[v];
head[v] = ecnt++; }
void bfs()
{
memset(dis + , 0x3f, n * sizeof(int));
queue<int> que;
que.push(ed);
dis[ed] = ;
while(!que.empty())
{
int u = que.front();
que.pop();
gap[dis[u]]++;
for(int p = head[u]; ~p; p = next[p])
{
int v = to[p];
if(flow[p ^ ] && dis[u] + < dis[v])
{
dis[v] = dis[u] + ;
que.push(v);
}
}
}
} int max_flow(int ss, int tt)
{
st = ss, ed = tt;
int ans = , minFlow = INF;
for(int i = ; i <= n; ++i)
{
cur[i] = head[i];
gap[i] = ; }
bfs();
int u = pre[st] = st;
while(dis[st] < n)
{
bool flag = false;
for(int &p = cur[u]; ~p; p = next[p])
{
int v = to[p];
if(flow[p] && dis[u] == dis[v] + )
{
flag = true;
minFlow = min(minFlow, flow[p]);
pre[v] = u;
u = v;
if(u == ed)
{
ans += minFlow;
while(u != st)
{
u = pre[u];
flow[cur[u]] -= minFlow;
flow[cur[u] ^ ] += minFlow; }
minFlow = INF; }
break; } }
if(flag) continue;
int minDis = n - ;
for(int p = head[u]; ~p; p = next[p])
{
int &v = to[p];
if(flow[p] && dis[v] < minDis)
{
minDis = dis[v];
cur[u] = p; }
}
if(--gap[dis[u]] == ) break;
++gap[dis[u] = minDis + ];
u = pre[u]; }
return ans; } int stk[MAXV], top;
bool sccno[MAXV], vis[MAXV];
bool dfs(int u, int f, bool flag)
{
vis[u] = true;
stk[top++] = u;
for(int p = head[u]; ~p; p = next[p]) if(flow[p])
{
int v = to[p];
if(v == f) continue;
if(!vis[v])
{
if(dfs(v, u, flow[p ^ ])) return true; }
else if(!sccno[v]) return true; }
if(!flag)
{
while(true)
{
int x = stk[--top];
sccno[x] = true;
if(x == u) break; } }
return false; }
bool acycle()
{
memset(sccno + , , n * sizeof(bool));
memset(vis + , , n * sizeof(bool));
top = ;
return dfs(ed, , ); }
} G;
int row[MAXN], col[MAXN];
int mat[MAXN][MAXN];
int n, m, k, ss, tt;
void solve()
{
int sumr = accumulate(row + , row + n + , );
int sumc = accumulate(col + , col + m + , );
if(sumr != sumc)
{
puts("Impossible");
return ; }
int res = G.max_flow(ss, tt);
if(res != sumc)
{
puts("Impossible");
return ; }
if(G.acycle())
{
puts("Not Unique"); }
else
{
puts("Unique");
for(int i = ; i <= n; ++i)
{
for(int j = ; j < m; ++j) printf("%d ", G.flow[mat[i][j]]);
printf("%d\n", G.flow[mat[i][m]]); } } }
int main()
{
while(scanf("%d%d%d", &n, &m, &k) != EOF)
{
for(int i = ; i <= n; ++i) scanf("%d", &row[i]);
for(int i = ; i <= m; ++i) scanf("%d", &col[i]);
ss = n + m + , tt = n + m + ;
G.init(tt);
for(int i = ; i <= n; ++i) G.add_edge(ss, i, row[i]);
for(int i = ; i <= m; ++i) G.add_edge(n + i, tt, col[i]);
for(int i = ; i <= n; ++i)
{
for(int j = ; j <= m; ++j)
{
mat[i][j] = G.ecnt ^ ;
G.add_edge(i, n + j, k); }
}
solve(); }
}
这类题目，这个作为模版哦！