[最大流唯一性判断]hdoj 4888

题意

 

给出一个矩阵n行每一行数字的和，m列每列数字的和，以及矩阵上单个数字的最大值k    N(1 ≤ N ≤ 400) , M(1 ≤ M ≤ 400) and K(1 ≤ K ≤ 40)。问是否存在这样的矩阵，存在的话，这个矩阵是否唯一，如果唯一输出这个矩阵。

 

思路

 

存在性的判断比较简单，求出最大流之后查看最大流是否等于所有数字的和即可。关键是唯一性，之前做过的最小割唯一性zoj 2587:Unique Attack 并不适用这道题，忍不住去看了题解，官方解释是“解唯一的充分必要条件是完成最大流后的残余网络没有长度大于 2 的环。”，也就是说，如果存在环的话，流量依然可以在环内流动，所以最大流不唯一。

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int inf=1<<28;
const int nMax=1000;
const int mMax=800000;

class node{
    public:
    int c,u,v,next;
};node edge[mMax];
int ne, head[nMax];
int cur[nMax], ps[nMax], dep[nMax];

void addedge(int u, int v,int w){   ////dinic
    edge[ne].u = u;
    edge[ne].v = v;
    edge[ne].c = w;
    edge[ne].next = head[u];
    head[u] = ne ++;
    edge[ne].u = v;
    edge[ne].v = u;
    edge[ne].c = 0;
    edge[ne].next = head[v];
    head[v] = ne ++;
}

int dinic(int s, int t){                       //  dinic
    int tr, res = 0;
    int i, j, k, f, r, top;
    while(1){
        memset(dep, -1, sizeof(dep));
        for(f = dep[ps[0]=s] = 0, r = 1; f != r;)
            for(i = ps[f ++], j = head[i]; j; j = edge[j].next)
                if(edge[j].c && dep[k=edge[j].v] == -1){
                    dep[k] = dep[i] + 1;
                    ps[r ++] = k;
                    if(k == t){
                        f = r; break;
                    }
                }
        if(dep[t] == -1) break;
        memcpy(cur, head, sizeof(cur));
        i = s, top = 0;
        while(1){
            if(i == t){
                for(tr =inf, k = 0; k < top; k ++)
                    if(edge[ps[k]].c < tr)
                        tr = edge[ps[f=k]].c;
                for(k = 0; k < top; k ++){
                    edge[ps[k]].c -= tr;
                    edge[ps[k]^1].c += tr;
                }
                i = edge[ps[top=f]].u;
                res += tr;
            }
            for(j = cur[i]; cur[i]; j = cur[i] =edge[cur[i]].next){
                if(edge[j].c && dep[i]+1 == dep[edge[j].v]) break;
            }
            if(cur[i]){
                ps[top ++] = cur[i];
                i = edge[cur[i]].v;
            }
            else{
                if(top == 0) break;
                dep[i] = -1;
                i = edge[ps[-- top]].u;
            }
        }
    }
    return res;
}

bool vis[nMax];
bool dfs(int loc ,int fa){
    for(int i=head[loc];i;i=edge[i].next){
        int v=edge[i].v;
       // if(v==fa||edge[i].c==0)continue;
        if(edge[i].c==0||i==(fa^1))continue;
        if(vis[v])return 1;
        vis[v]=1;
        if(dfs(v,i))return 1;
        vis[v]=0;
    }
    return 0;
}

int aaa[500],bbb[500],ans[500][500];
int main(){
    int n,m,k,i,j,sum1,sum2;
    while(cin>>n>>m>>k){
        sum1=sum2=0;
        for(i=1;i<=n;i++){
            cin>>aaa[i];
            sum1+=aaa[i];
        }
        for(i=1;i<=m;i++){
            cin>>bbb[i];
            sum2+=bbb[i];
        }
        if(sum1!=sum2){
            cout<<"Impossible\n";
            continue;
        }
        ne=2;
        memset(head,0,sizeof(head));
        for(i=1;i<=n;i++){
            addedge(0,i,aaa[i]);
            for(j=n+1;j<=n+m;j++){
                addedge(i,j,k);
            }
        }
        for(i=n+1;i<=n+m;i++){
            addedge(i,n+m+1,bbb[i-n]);
        }
        sum2=dinic(0,n+m+1);

        if(sum1!=sum2){
            cout<<"Impossible\n";
            continue;
        }

        bool flag=0;
        memset(vis,0,sizeof(vis));
        for(i=1;i<=n;i++){
            if(dfs(i,-1)){
                flag=1;
                break;
            }
        }
        if(flag){
            cout<<"Not Unique"<<endl;
        }else{
            cout<<"Unique"<<endl;
            for(i=1;i<=n;i++){
                for(j=head[i];j;j=edge[j].next){
                    int v=edge[j].v;
                    if(v<=n||v>n+m)continue;
                    ans[i][v-n]=k-edge[j].c;
                }
            }
            for(i=1;i<=n;i++){
                for(j=1;j<m;j++){
                    cout<<ans[i][j]<<" ";
                }cout<<ans[i][j]<<endl;
            }
        }
    }
    return 0;
}