Hdoj 4888 Redraw Beautiful drawings "Maximum flow full stream + uniqueness judgment"

Title: Hdoj 4888 Redraw Beautiful drawings

Category: Maximum flow full flow, maximum flow uniqueness

Source: multi-university Training Contest 3

Test instructions: A matrix of each row of each column and all know, and then let you beg can fill, whether the only, the only words output solution,

Analysis: This topic can be seen as the maximum flow, but the difficulties are 2.

First of all: in the Matrix 400 * 400, fortunately, the matrix and no special requirements, so you can directly to the row and column to build a plan:

1. Source Point  - each row corresponds to a point where the traffic is limited to the row's and

2. each row corresponds to a point  - Each column corresponds to a point where the flow limit is K

3. Each column corresponds to a point  - meeting Point, the flow limit for the column and the

Time Complexity O (n^2* m) with dinci words

The 2nd is the sole judgment of the solution:

The

is given in the following: The only sufficient condition for the solution is to complete the maximum flow after the residual network is not longer than the   2  

A bit confused, but we can use the violent law to judge, that is, if J is not full stream in the same line, and K is not an empty stream, and the other row exists K is not full stream, and J is not an empty stream, think the same column is only limited, then we can give J is not empty stream minus a certain value, not the same Then k the same operation, such rows and columns and also satisfies the conditions.

Based on the above facts, we can hold a column to judge the time Complexity O (n^3)

Down Total time complexity O (n^2* (m+n)) n for 400,m = n^2, down about n^4 = 256 0000 0000 (seems a bit large)

After submission g++, C + + more than 1000, C + + unexpectedly, a bit abnormal.

This topic with EK words will time out, I began to write with Ek Super, because the EK complexity O (n*m^2) about N^5

AC Code:

#include <cstdio> #include <cstring> #include <iostream> #include <string> #include < algorithm> #include <vector> #include <queue>using namespace std; #define Del (b) memset (A,b,sizeof (a)) const int N = 1000;const int inf = 0x3f3f3f3f;int n,m,k;struct node{int from,to,cap,flow;};  Vector<int> v[n];vector<node> E;int Vis[n];    Build the hierarchy diagram int cur[n];void add_node (int from,int to,int cap) {E.push_back ((Node) {from,to,cap,0});    E.push_back (Node) {to,from,0,0});    int tmp=e.size ();    V[from].push_back (tmp-2); V[to].push_back (tmp-1);}    BOOL BFs (int s,int t) {Del (vis,-1);    Queue<int> Q;    Q.push (s);    Vis[s] = 0;        while (!q.empty ()) {int X=q.front ();        Q.pop ();            for (int i=0; i<v[x].size (); i++) {Node TMP = e[v[x][i]];                if (vis[tmp.to]<0 && tmp.cap>tmp.flow)//The second condition guarantees {vis[tmp.to]=vis[x]+1;            Q.push (tmp.to);   }     }} if (vis[t]>0) return true; return false;}    int dfs (int o,int f,int t) {if (o==t | | f==0)//optimize return F;    int a = 0,ans=0;        for (int &i=cur[o]; i<v[o].size (); i++)//Note front ' & ', very important optimization {Node &tmp = e[v[o][i]];            if (vis[tmp.to]== (vis[o]+1) && (a = DFS (Tmp.to,min (f,tmp.cap-tmp.flow), T)) >0) {tmp.flow+=a; E[v[o][i]^1].flow-=a;            Ans+=a of the mode of storage and mapping;            F-=a;        if (f==0)//attention optimization break;  }} return ans;    Optimized}int dinci (int s,int t) {int ans=0;        while (BFS (s,t)) {Del (cur,0);        int Tm=dfs (S,INF,T);    Ans+=tm; } return ans;        int mp[450][450];int Dou[450][450];bool Solve () {for (int i=1;i<=n;i++) {int tmp=0;    for (int j=0;j<v[i].size (); j + +)//Note standard if (e[v[i][j]].to>n) Mp[i][++tmp]=e[v[i][j]].flow;    } Del (dou,0); for (int i = 1; I <= n; i++)//violence method according to full flow empty flowJudgment {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;}        int main () {while (~scanf ("%d%d%d", &n,&m,&k)) {int x;        int s=0,t=m+n+1,sum_a=0,sum_b=0;            for (int i=1;i<=n;i++) {scanf ("%d", &x);            Sum_a + = x;            Add_node (S,I,X);        for (int j=1;j<=m;j++) Add_node (i,n+j,k);            } for (int i=1;i<=m;i++) {scanf ("%d", &x);            Sum_b+=x; Add_node (i+n, t,x);        } int Ans=dinci (S,T);        if (Ans!=min (Sum_a,sum_b)) printf ("impossible\n");            else {memset (mp,0,sizeof (MP));            if (Solve () ==0) printf ("Not unique\n");                else {printf ("unique\n"); for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) printf ("%d%c", mp[i][j],j==m? '            \ n ': ');        }} for (int i=0; i<=t; i++) v[i].clear ();    E.clear (); } return 0;}

Hdoj 4888 Redraw Beautiful drawings "Maximum flow full stream + uniqueness judgment"