[网络流最大流经典][uva 11082][矩阵解压]

题目大意

这里写图片描述

分析

这里写图片描述

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
using namespace std;
const int MAXN=200+5;
const int MAXM=2000;
const int INF=0x3f3f3f3f;
struct Edge
{
    int to,next,cap,flow;
    void get(int a,int b,int c,int d)
    {
        to=a;next=b;cap=c;flow=d;
    }
}edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
}
//单向图三个参数,无向图四个参数
void addedge(int u,int v,int w,int rw=0)
{
    edge[tol].get(v,head[u],w,0);head[u]=tol++;
    edge[tol].get(u,head[v],rw,0);head[v]=tol++;
}
int sap(int start,int end,int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u=start;
    pre[u]=-1;
    gap[0]=N;
    int ans=0;
    while(dep[start]<N)
    {
        if(u==end)
        {
            int Min=INF;
            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])
                if(Min>edge[i].cap-edge[i].flow)
                   Min=edge[i].cap-edge[i].flow;
            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])
            {
                edge[i].flow+=Min;
                edge[i^1].flow-=Min;
            }
            u = start;
            ans+=Min;
            continue;
        }
        bool flag=false;
        int v;
        for(int i=cur[u];i !=-1;i=edge[i].next)
        {
            v=edge[i].to;
            if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u])
            {
                flag=true;
                cur[u]=pre[v]=i;
                break;
            }
        }
        if(flag)
        {
            u=v;
            continue;
        }
        int Min=N;
        for(int i=head[u];i!=-1;i=edge[i].next)
            if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min)
        {
            Min=dep[edge[i].to];
            cur[u]=i;
        }
        gap[dep[u]]--;
        if(!gap[dep[u]]) return ans;
        dep[u]=Min+1;
        gap[dep[u]]++;
        if(u!=start) u=edge[pre[u]^1].to;
    }
    return ans;
}
int R,C;
int A[50],B[50];
int ANS[50][50];
int sum=0; 
void input()
{
    sum=0;
    init();
    cin>>R>>C;
    for(int i=1;i<=R;i++)
    scanf("%d",&A[i]);
    for(int i=1;i<=C;i++)
    scanf("%d",&B[i]);
    for(int i=R;i>=1;i--)
    {
    A[i]=A[i]-A[i-1];
    sum=sum+A[i];
    }
    for(int i=C;i>=1;i--)
    B[i]=B[i]-B[i-1];
}
void solve()
{
    int S=0,T=R+C+1;
    //S->
    for(int i=1;i<=R;i++)
    {
        addedge(S,i,A[i]-C);
    }
    //->T 
    for(int i=1;i<=C;i++)
    {
        addedge(R+i,T,B[i]-R);
    }
    for(int i=1;i<=R;i++)
        for(int j=1;j<=C;j++)
        {
            addedge(i,R+j,19);
        }
}
int main()
{
//  freopen("a.in","r",stdin);
    int T;
    cin>>T;
    int CASE=0;
    while(T--)
    {
        printf("Matrix %d
",++CASE);
        input();
        solve();
        int k=sap(0,R+C+1,R+C+2);
        for(int i=1;i<=R;i++)
        {
            for(int p=head[i];p!=-1;p=edge[p].next)
            {
                ANS[i][edge[p].to-R]=edge[p].flow+1; 
            }
        }
        for(int i=1;i<=R;i++)
            {
                for(int j=1;j<=C;j++)
                {
                    printf("%d",ANS[i][j]);
                    if(j!=C) printf(" ");
                }
                printf("
");
            } 
        printf("
"); 
    }

}
原文地址:https://www.cnblogs.com/zy691357966/p/5480305.html