HDU3605Escape（最大流SAP+状态压缩优化点的个数）

Escape

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6239    Accepted Submission(s): 1474

Problem Description

2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.

 

Input

More set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) n indicate there n people on the earth, m representatives m planet, planet and people labels are from 0. Here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000

 

Output

Determine whether all people can live up to these stars
If you can output YES, otherwise output NO.

 

Sample Input

1 1 1 1 2 2 1 0 1 0 1 1

 

Sample Output

YES NO

 

Source

2010 ACM-ICPC Multi-University Training Contest（17）——Host by ZSTU

 

题意：现在地球有n（<=100w）人数要去m（<=10）个其他星球生存，接下来n行每行m列，如果第i 为 1 对应这个人可去星球 i ，第 i 个星球最多可容ai个人（最后一行给出m个数）。现在问是否所有的人都可以到其他星球生存。

解题：最大流SAP。建图：因为人数多，而去星球的状态少最多是1024种，所以可以把每种状态看作是点，源点与每种状态连接一条边，容量为当前状态的人数，再每种状态与相应的星球连边，边容为当前状态的人数。再每个星球与汇点连边，边容为星球能容纳的人数。

另一种方法解：多重匹配，人作x部分和星球作y部分，形成二部图。

下面是最大流SAP解法：用G++  。

#include
#include
#include
#include
using namespace std;
#define captype int

const int MAXN = 1030;   //点的总数
const int MAXM = 14000;    //边的总数
const int INF = 1<<30;

struct EDG{
    int to,next;
    captype cap,flow;
} edg[MAXM];
int eid,head[MAXN];
int gap[MAXN];  //每种距离(或可认为是高度)点的个数
int dis[MAXN];  //每个点到终点eNode 的最短距离
int cur[MAXN];  //cur[u] 表示从u点出发可流经 cur[u] 号边
int pre[MAXN];

inline void init(){
    eid=0;
    memset(head,-1,sizeof(head));
}
//有向边 三个参数，无向边4个参数
inline void addEdg(int u,int v,captype c,captype rc=0){
    edg[eid].to=v; edg[eid].next=head[u];
    edg[eid].cap=c; edg[eid].flow=0; head[u]=eid++;

    edg[eid].to=u; edg[eid].next=head[v];
    edg[eid].cap=rc; edg[eid].flow=0; head[v]=eid++;
}
inline captype maxFlow_sap(int& sNode,int& eNode, int n){//n是包括源点和汇点的总点个数，这个一定要注意
    memset(gap,0,sizeof(gap));
    memset(dis,0,sizeof(dis));
    memcpy(cur,head,sizeof(head));
    pre[sNode] = -1;
    gap[0]=n;
    captype ans=0;  //最大流
    int u=sNode  , i ;
    while(dis[sNode]edg[i].cap-edg[i].flow){
                    Min=edg[i].cap-edg[i].flow;
                    inser=i;
                }
                i=pre[edg[i^1].to];
            }

            i=pre[u];
            while( i!=-1 ){
                edg[i].flow+=Min;
                edg[i^1].flow-=Min;  //可回流的边的流量
                i=pre[edg[i^1].to];
            }
            ans+=Min;
            u=edg[inser^1].to;
            continue;
        }
        bool flag = false;  //判断能否从u点出发可往相邻点流
        int v;
        i=cur[u];
        while(  i!=-1 ){
            v=edg[i].to;
            if(edg[i].cap-edg[i].flow>0 && dis[u]==dis[v]+1){
                flag=true;
                cur[u]=pre[v]=i;
                break;
            }
            i=edg[i].next;
        }
        if(flag){
            u=v;
            continue;
        }
        //如果上面没有找到一个可流的相邻点，则改变出发点u的距离（也可认为是高度）为相邻可流点的最小距离+1
        int Mind= n;
        i=head[u];
        while(  i!=-1 ){
             if(edg[i].cap-edg[i].flow>0 && Mind>dis[edg[i].to]){
                 Mind=dis[edg[i].to];
                 cur[u]=i;
             }
            i=edg[i].next;
        }

        gap[dis[u]]--;
        if(gap[dis[u]]==0) return ans;  //当dis[u]这种距离的点没有了，也就不可能从源点出发找到一条增广流路径
                                        //因为汇点到当前点的距离只有一种，那么从源点到汇点必然经过当前点，然而当前点又没能找到可流向的点，那么必然断流
        dis[u]=Mind+1;//如果找到一个可流的相邻点，则距离为相邻点距离+1，如果找不到，则为n+1
        gap[dis[u]]++;
        if(u!=sNode) u=edg[pre[u]^1].to;  //退一条边
    }
    return ans;
}
inline void scanf(int& valu){
    char ch;
    while(ch=getchar()){
        if(ch>='0'&&ch<='9')
            break;
    }
    valu=ch-'0';
    while(ch=getchar()){
        if(ch<'0' || ch>'9')
            break;
        valu=valu*10+ch-'0';
    }
}

int main(){
    int n,m,a , s ,t ,ans ,k[1030] ;
    while(scanf("%d%d",&n,&m)>0){
        memset(k,0,sizeof(k));
        int id=0 , i=0 ,j;
        while(i0 ){
            if(k[i]>0){ //出现的状态
                addEdg(s , id , k[i]);
                j=0;
                while( (1<0?:);
    }
}