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Redraw Beautiful Drawings

Time Limit: 3000/1500 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1660????Accepted Submission(s): 357


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.

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).
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
Author Fudan University
給你一個n*m的矩陣，然后個格子里面有一個小于k的數，而且告訴你每行和每列的和，讓你求是否存在解，并推斷是否唯一。 建圖：源點和每行連邊，容量為每行的和。每列和匯點連邊，容量為每列的和。每行和每列連邊，容量為k。 推斷是否存在唯一解，在殘留網絡里面是否能找到一個環。 //515MS 7304K #include<stdio.h> #include<string.h> #define M 1007 int s,t,n,m,k,sum1,sum2; int row[M],col[M],vis[M],g[M][M]; const int MAXN=20010;//點數的最大值 const int MAXM=880010;//邊數的最大值 const int INF=0x3f3f3f3f; struct Node {int from,to,next;int cap; }edge[MAXM]; int tol; int head[MAXN]; int dis[MAXN]; int gap[MAXN];//gap[x]=y :說明殘留網絡中dis[i]==x的個數為y void init() {tol=0;memset(head,-1,sizeof(head)); } void addedge(int u,int v,int w) {edge[tol].from=u;edge[tol].to=v;edge[tol].cap=w;edge[tol].next=head[u];head[u]=tol++;edge[tol].from=v;edge[tol].to=u;edge[tol].cap=0;edge[tol].next=head[v];head[v]=tol++; } void BFS(int start,int end) {memset(dis,-1,sizeof(dis));memset(gap,0,sizeof(gap));gap[0]=1;int que[MAXN];int front,rear;front=rear=0;dis[end]=0;que[rear++]=end;while(front!=rear){int u=que[front++];if(front==MAXN)front=0;for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].to;if(dis[v]!=-1)continue;que[rear++]=v;if(rear==MAXN)rear=0;dis[v]=dis[u]+1;++gap[dis[v]];}} } int SAP(int start,int end) {int res=0,nn=end+1;BFS(start,end);int cur[MAXN];int S[MAXN];int top=0;memcpy(cur,head,sizeof(head));int u=start;int i;while(dis[start]<nn){if(u==end){int temp=INF;int inser;for(i=0;i<top;i++)if(temp>edge[S[i]].cap){temp=edge[S[i]].cap;inser=i;}for(i=0;i<top;i++){edge[S[i]].cap-=temp;edge[S[i]^1].cap+=temp;}res+=temp;top=inser;u=edge[S[top]].from;}if(u!=end&&gap[dis[u]-1]==0)//出現斷層，無增廣路break;for(i=cur[u];i!=-1;i=edge[i].next)if(edge[i].cap!=0&&dis[u]==dis[edge[i].to]+1)break;if(i!=-1){cur[u]=i;S[top++]=i;u=edge[i].to;}else{int min=nn;for(i=head[u];i!=-1;i=edge[i].next){if(edge[i].cap==0)continue;if(min>dis[edge[i].to]){min=dis[edge[i].to];cur[u]=i;}}--gap[dis[u]];dis[u]=min+1;++gap[dis[u]];if(u!=start)u=edge[S[--top]].from;}}return res; }void build() {sum1=0,sum2=0;for(int i=1;i<=n;i++){scanf("%d",&row[i]);sum1+=row[i];addedge(s,i,row[i]);}for(int i=1;i<=m;i++){scanf("%d",&col[i]);sum2+=col[i];addedge(n+i,t,col[i]);}for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)addedge(i,n+j,k); }bool iscycle(int u,int fa) {for(int i=head[u];i!=-1;i=edge[i].next){if(i==(fa^1))continue;if(edge[i].cap){if(vis[edge[i].to])return true;vis[edge[i].to]=true;if(iscycle(edge[i].to,i))return true;vis[edge[i].to]=false;}}return false; } void solve() {if(sum1!=sum2){printf("Impossible\n");return;}int anss=SAP(s,t);//printf("anss=%d\n",anss);if(anss!=sum1){printf("Impossible\n");return;}memset(vis,false,sizeof(vis));int flag=0;for(int i=1;i<=n;i++)//推斷殘留網絡是否存在環if(iscycle(i,-1)){flag=1;break;}if(flag)printf("Not Unique\n");else{printf("Unique\n");for(int u=1;u<=n;u++)for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].to;if(v>n&&v<=n+m)g[u][v-n]=k-edge[i].cap;}for(int i=1;i<=n;i++){for(int j=1;j<m;j++)printf("%d ",g[i][j]);printf("%d\n",g[i][m]);}} } int main() {while(scanf("%d%d%d",&n,&m,&k)!=EOF){s=0,t=n+m+1;init();build();solve();} }

轉載于:https://www.cnblogs.com/mengfanrong/p/4295453.html

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