网络最大流解方程组
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多校联赛第三场（hdu4888）
Redraw Beautiful Drawings
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1320 Accepted Submission(s): 268
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.
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.
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 44 24 2 4 2 22 2 5 05 41 4 391 2 3 3
Sample Output
Not UniqueImpossibleUnique1 2 3 3
建图：建立二部图，把行标列为x部，把列表列为Y部，设立源点source和汇点sink，把行标和源点建边，流量是行和，把列标和汇点建边流量是列和，然后把行标和列标两两相连，代表第i行第j列的数，流量为k，跑完一次最大流（Dinic或Isap）如果最大流量不等于所有数的总和，则无解，如果存在残留网络中存在一个正环（顶点个数>2且环的残余流量>0）则为多解，否则为唯一解，然后遍历遍历完所有的行标i以及行标j对应的列标，则<i,j>所代表的反向边的流量就是坐标为（i，j）的值，存储记录输出即可；
程序;
#include"stdio.h"#include"string.h"#include"iostream"#include"map"#include"string"#include"queue"#include"stdlib.h"#include"math.h"#define M 200000#define N 10000#define eps 1e-10#define inf 1000000000#define mod 2333333using namespace std;struct node{ int u,v,w,next;}edge[M*2];int t,head[N],row[N],col[N],q[N],dis[N],work[N],use[N],flag;void init(){ t=0; memset(head,-1,sizeof(head));}void add(int u,int v,int w){ edge[t].u=u; edge[t].v=v; edge[t].w=w; edge[t].next=head[u]; head[u]=t++; edge[t].u=v; edge[t].v=u; edge[t].w=0; edge[t].next=head[v]; head[v]=t++;}int bfs(int S,int T){ int rear=0; memset(dis,-1,sizeof(dis)); dis[S]=0; q[rear++]=S; for(int i=0;i<rear;i++) { for(int j=head[q[i]];j!=-1;j=edge[j].next) { int v=edge[j].v; if(edge[j].w&&dis[v]==-1) { dis[v]=dis[q[i]]+1; q[rear++]=v; if(v==T) return 1; } } } return 0;}int dfs(int cur,int a,int T){ if(cur==T) return a; for(int &i=work[cur];i!=-1;i=edge[i].next) { int v=edge[i].v; if(edge[i].w&&dis[v]==dis[cur]+1) { int tt=dfs(v,min(a,edge[i].w),T); if(tt) { edge[i].w-=tt; edge[i^1].w+=tt; return tt; } } } return 0;}int Dinic(int S,int T){ int ans=0; while(bfs(S,T)) { memcpy(work,head,sizeof(head)); while(int tt=dfs(S,inf,T)) ans+=tt; } return ans;}int mp[444][444];int DFS(int u,int f){ use[u]=1; for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].v; if(edge[i].w&&v!=f) { if(use[v]) return 1; if(DFS(v,u)) return 1; } } use[u]=0; return 0;}int judge(int n,int m){ //printf("%d\n",DFS(1,1)); for(int i=1;i<=n;i++) { memset(use,0,sizeof(use)); if(DFS(i,i)) return 1; } return 0;}int main(){ int n,m,k,i,j; while(scanf("%d%d%d",&n,&m,&k)!=-1) { int r=0,c=0; for(i=1;i<=n;i++) { scanf("%d",&row[i]); r+=row[i]; } for(j=1;j<=m;j++) { scanf("%d",&col[j]); c+=col[j]; } if(r!=c) { printf("Impossible\n"); continue; } init(); for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { add(i,n+j,k); } } for(i=1;i<=n;i++) add(0,i,row[i]); for(j=1;j<=m;j++) add(j+n,m+n+1,col[j]); int ans=Dinic(0,m+n+1); if(ans<r) { printf("Impossible\n"); continue; } if(judge(n,m)) { printf("Not Unique\n"); continue; } printf("Unique\n"); t=0; for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { if(j==1) printf("%d",edge[t^1].w); else printf(" %d",edge[t^1].w); t=t+2; } printf("\n"); } } return 0;}</pre><pre code_snippet_id="438794" snippet_file_name="blog_20140730_1_6325119" name="code" class="cpp">加矩阵DP优化判是否唯一：时间171ms
#include"stdio.h"#include"string.h"#include"iostream"#include"map"#include"string"#include"queue"#include"stdlib.h"#include"math.h"#define M 800000#define N 1900#define eps 1e-10#define inf 1000000000#define mod 2333333using namespace std;struct node{ int u,v,w,next;}edge[M];int t,head[N],row[N],col[N],q[N],dis[N],work[N],use[N],flag;void init(){ t=0; memset(head,-1,sizeof(head));}void add(int u,int v,int w){ edge[t].u=u; edge[t].v=v; edge[t].w=w; edge[t].next=head[u]; head[u]=t++; edge[t].u=v; edge[t].v=u; edge[t].w=0; edge[t].next=head[v]; head[v]=t++;}int bfs(int S,int T){ int rear=0; memset(dis,-1,sizeof(dis)); dis[S]=0; q[rear++]=S; for(int i=0;i<rear;i++) { for(int j=head[q[i]];j!=-1;j=edge[j].next) { int v=edge[j].v; if(edge[j].w&&dis[v]==-1) { dis[v]=dis[q[i]]+1; q[rear++]=v; if(v==T) return 1; } } } return 0;}int dfs(int cur,int a,int T){ if(cur==T) return a; for(int &i=work[cur];i!=-1;i=edge[i].next) { int v=edge[i].v; if(edge[i].w&&dis[v]==dis[cur]+1) { int tt=dfs(v,min(a,edge[i].w),T); if(tt) { edge[i].w-=tt; edge[i^1].w+=tt; return tt; } } } return 0;}int Dinic(int S,int T){ int ans=0; while(bfs(S,T)) { memcpy(work,head,sizeof(head)); while(int tt=dfs(S,inf,T)) ans+=tt; } return ans;}int mp[555][555];int G[555][555];int judge(int n,int m,int k){ int i,j,r; t=0; for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { G[i][j]=edge[t^1].w; t+=2; } } memset(mp,0,sizeof(mp)); for(i=1;i<=n;i++) { if(row[i]==0||row[i]==k*m)continue; for(j=1;j<=m;j++) { if(col[j]==0||col[j]==k*n)continue; for(r=j+1;r<=m;r++) { int f1=0,f2=0; if(G[i][j]<k&&G[i][r]>0) { if(mp[r][j]) return 1; f1++; } if(G[i][j]>0&&G[i][r]<k) { if(mp[j][r]) return 1; f2++; } if(f1)mp[j][r]=1; if(f2)mp[r][j]=1; } } } return 0;}int main(){ int n,m,k,i,j; while(scanf("%d%d%d",&n,&m,&k)!=-1) { int r=0,c=0; for(i=1;i<=n;i++) { scanf("%d",&row[i]); r+=row[i]; } for(j=1;j<=m;j++) { scanf("%d",&col[j]); c+=col[j]; } if(r!=c) { printf("Impossible\n"); continue; } int flag=0; for(i=1;i<=n;i++) if(m*k<row[i]) flag++; for(i=1;i<=m;i++) if(n*k<col[i]) flag++; if(flag) { printf("Impossible\n"); continue; } init(); for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { add(i,n+j,k); } } for(i=1;i<=n;i++) add(0,i,row[i]); for(j=1;j<=m;j++) add(j+n,m+n+1,col[j]); int ans=Dinic(0,m+n+1); if(ans<r) { printf("Impossible\n"); continue; } if(judge(n,m,k)) { printf("Not Unique\n"); continue; } printf("Unique\n"); t=0; for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { if(j==1) printf("%d",edge[t^1].w); else printf(" %d",edge[t^1].w); t=t+2; } printf("\n"); } } return 0;} 0 0
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