poj2391 二分最大流+拆点

来源：互联网发布：17年华研网络词作大赛编辑：程序博客网时间：2024/06/05 09:18

这道题范围开的非常的憋屈人

首先建立一个超级源点编号是0 一个超级汇点编号是2*f+1

将每个避雨点分成两个点第一组编号为1～f 对应的第二组为f+1～2*f

每次建图从源点向第一组每个避雨点连边边的值为现在停留在这个点的牛的数量

从第二组每个点向汇点进行连边边的值为这个点可容纳的牛的数量

然后对时间进行二分如果从避雨点i走到避雨点j的时间小于等于mid

那么就从i到j+f建一条边边的值为INF 这样如果某个点牛的数量超过了容纳量这个点的牛就可以流到其他的点了

然后从0到2*f+1跑最大流如果最大流>=牛的总数量那么mid有减小的余地最后输出ans即可

#include<algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<string>#include<stack>#include<queue>#include<cmath>#include<stack>#include<list>#include<set>typedef long long ll;using namespace std;const int MAXN=500;//jiedian de zui da zhiconst int MAXM=505000;//bian de zui da zhiconst int INF=0xFFFFFFF;const ll MAX=0x7FFFFFFFFFFFFFF;int bian[1505][2];ll map[205][205];struct Node{    int from,to,next;    int cap;}edge[MAXM];int tol;int head[MAXN];int dep[MAXN];int gap[MAXN];void init()  //remember write it in main function{    tol=0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int w){    edge[tol].from=u;    edge[tol].to=v;    edge[tol].cap=w;    edge[tol].next=head[u];    head[u]=tol++;    edge[tol].from=v;    edge[tol].to=u;    edge[tol].cap=0;//wuxiangtu  this place change to w;    edge[tol].next=head[v];    head[v]=tol++;}void BFS(int start,int end){    memset(dep,-1,sizeof(dep));    memset(gap,0,sizeof(gap));    gap[0]=1;    int que[MAXN];    int front,rear;    front=rear=0;    dep[end]=0;    que[rear++]=end;    while(front!=rear)    {        int u=que[front++];        if(front==MAXN)front=0;        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].to;            if(dep[v]!=-1)continue;            que[rear++]=v;            if(rear==MAXN)rear=0;            dep[v]=dep[u]+1;            ++gap[dep[v]];        }    }}int SAP(int start,int end,int n) //n shi jiedian de zui da ge shu ,including source and sink{    int res=0;    BFS(start,end);    int cur[MAXN];    int S[MAXN];    int top=0;    memcpy(cur,head,sizeof(head));    int u=start;    int i;    while(dep[start]<n)    {        if(u==end)        {            int temp=INF;            int inser;            for(i=0;i<top;i++)                if(temp>edge[S[i]].cap)                {                    temp=edge[S[i]].cap;                    inser=i;                }            for(i=0;i<top;i++)            {                edge[S[i]].cap-=temp;                edge[S[i]^1].cap+=temp;            }            res+=temp;            top=inser;            u=edge[S[top]].from;        }        if(u!=end&&gap[dep[u]-1]==0)//≥ˆœ÷∂œ≤„£¨Œﬁ‘ˆπ„¬∑            break;        for(i=cur[u];i!=-1;i=edge[i].next)            if(edge[i].cap!=0&&dep[u]==dep[edge[i].to]+1)                break;        if(i!=-1)        {            cur[u]=i;            S[top++]=i;            u=edge[i].to;        }        else        {            int min=n;            for(i=head[u];i!=-1;i=edge[i].next)            {                if(edge[i].cap==0)continue;                if(min>dep[edge[i].to])                {                    min=dep[edge[i].to];                    cur[u]=i;                }            }            --gap[dep[u]];            dep[u]=min+1;            ++gap[dep[u]];            if(u!=start)u=edge[S[--top]].from;        }    }    return res;}void floyd(int n){            for(int i=1; i<=n; i++)        for(int j=1; j<=n; j++)            if (MAX != map[j][i])                for(int k=1; k<=n; k++)                    if (map[j][i] + map[i][k] < map[j][k])                        map[j][k] = map[j][i] + map[i][k];    }void getmap(int f,ll mid){    init();    int i,j;    for(i=1;i<=f;i++)    {        addedge(0,i,bian[i][0]);        addedge(i+f, 2*f+1, bian[i][1]);    }    for(i=1;i<=f;i++)    {        for(j=1;j<=f;j++)        {            if(map[i][j]<=mid)            {                addedge(i, f+j, INF);            }        }    }}int main(){    int i,j,f,p;    while(scanf("%d%d",&f,&p)==2)    {                for(i=1;i<=f;i++)        {            for(j=1;j<=f;j++)            {                    map[i][j]=MAX;            }            map[i][i]=0;                    }        int sum=0;                for(i=1;i<=f;i++)        {            scanf("%d%d",&bian[i][0],&bian[i][1]);            sum+=bian[i][0];                    }        //The first integer is the number of cows in that field. The second integer is the number of cows the shelter in that field can hold        ll x,y,z;        for(i=1;i<=p;i++)        {            scanf("%lld%lld%lld",&x,&y,&z);                map[x][y]=map[y][x]=min(map[x][y],z);        }        floyd(f);        ll l=0,r=MAX;        ll mid;        ll ans=-1;        while(l<r)        {            mid=(l+r)/2;            getmap(f,mid);            int tt=SAP(0, 2*f+1, 2*f+2);            if(tt>=sum)            {                r=mid;                ans=mid;                            }            else{                l=mid+1;            }        }        printf("%lld\n",ans);    }    return 0;}

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