[网络流最大流经典][uva 11082][矩阵解压]

来源：互联网发布：freecad for mac 编辑：程序博客网时间：2024/05/22 04:49
题目大意

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分析

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#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <ctime>#include <algorithm>#include <iostream>#include <sstream>#include <string>#define oo 0x13131313using 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\n",++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("\n");            }         printf("\n");     }}
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