【CodeChef-SPCLN】Cleaning the Space

https://odzkskevi.qnssl.com/7dfb262544887eff6fb35bfb444759d6?v=1502084197

 

做法是类似于最大割之类的东西，把每个碎片按照按钮拆点，从s到每个碎片第一个点连inf，每个碎片第i个按钮向第i+1个按钮连100-i能量。

然后最大流最小割，最大流=割掉的边和=100*n-能量和的最大值。

 

#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define dow(i,l,r) for(int i=r;i>=l;i--)
#define rep0(i,r) for(int i=0;i<r;i++)
#define repedge(i,x) for(int i=cur[x];i>=0;i=e[i].next)
#define maxn 20020
#define maxm 200100
#define LL long long
using namespace std;

typedef struct {
    int next,f,t;
}E;
E e[maxm];

int gap[maxn],d[maxn],fi[maxn],cur[maxn],s,t,total=0;
const int inf=100000000;
int n,m,q;

void add(int j,int k,int l)
{
    e[total].f=l;
    e[total].t=k;
    e[total].next=fi[j];
    fi[j]=total;
    total++;
}

void addedge(int j,int k,int l)
{
    add(j,k,l);
    add(k,j,0);
}


int sap(int x,int flow)
{
    if (x==t) return flow;
    int now=0;
    repedge(i,x) {
        int too=e[i].t;
        if (d[too]+1==d[x] && e[i].f) {
            int more=sap(too,min(e[i].f,flow-now));
            e[i].f-=more;
            e[i^1].f+=more;
            cur[x]=i;
            if (flow==(now+=more)) return flow;
        }
    }
    if (!(--gap[d[x]])) d[s]=t;
    gap[++d[x]]++;
    cur[x]=fi[x];
    return now;
}

int maxflow()
{
    memset(d,0,sizeof(d));
    memset(gap,0,sizeof(gap));
    rep(i,1,t) cur[i]=fi[i];
    gap[0]=t;
    int ans=0;
    while (d[s]<t) ans+=sap(s,inf);
    //printf("%d\n",ans);
    return ans;
}

int main()
{
  //  freopen("1.in","r",stdin);
    scanf("%d %d",&n,&m);
    memset(fi,-1,sizeof(fi));
    s=n*(m+1)+1;
    t=s+1;
    rep(i,1,n) {
        int now=(i-1)*(m+1);
        addedge(s,now+1,inf);
        addedge(now+m+1,t,inf);
        rep(j,1,m) {
            int k;
            scanf("%d",&k);
            if (k!=-1) addedge(now+j,now+j+1,100-k);
            else addedge(now+j,now+j+1,inf);
        }
    }
    scanf("%d",&q);
    while (q--) {
        int j,k;
        scanf("%d %d",&j,&k);
        rep(i,1,m) {
            int now1=(j-1)*(m+1)+i;
            int now2=(k-1)*(m+1)+i+1;
            addedge(now1,now2,inf);
        }
    }
    //printf("%d\n",maxflow());
    printf("%.2lf\n",(100.0*n-maxflow())/(n+0.0));
    return 0;
}

 

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