UVA 11419 SAM I AM (二分图，最小割)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2414

Problem C
SAM I AM
Input: Standard Input

Output: Standard Output

 

The world is in great danger!! Mental's forces have returned to Earth to eradicate humankind. Our last hope to stop this great evil is Sam “Serious” Stone. Equipped with various powerful weapons, Serious Sam starts his mission to destroy the forces of evil.

After fighting two days and three nights, Sam is now in front of the temple KOPTOS where Mental's general Ugh Zan III is waiting for him. But this time, he has a serious problem. He is in shortage of ammo and a lot of enemies crawling inside the temple waiting for him. After rounding thetemple Sam finds that the temple is in rectangle shape and he has the locations of all enemies in the temple.

All of a sudden he realizes that he can kill the enemies without entering the temple using the great cannon ball which spits out a gigantic ball bigger than him killing anything it runs into and keeps on rolling until it finally explodes. But the cannonball can only shoot horizontally or vertically and all the enemies along the path of that cannon ball will be killed.

Now he wants to save as many cannon balls as possible for fighting with Mental. So, he wants to know the minimum number of cannon balls and the positions from which he can shoot the cannonballs to eliminate all enemies from outside that temple.

 

Input

Here, the temple is defined as a RXC grid. The first line of each test case contains 3 integers: R(0<R<1001), C(0<C<1001) representing the grid of temple (R means number of row and C means number of column of the grid) and the number of enemies N(0<N<1000001) inside the temple. After that there are N lines each of which contains 2 integers representing the position of the enemies in that temple. Each test case is followed by a new line (except the last one). Input is terminated when R=C=N=0. The size of the input file is around 1.3 MB.

 

Output

For each test case there will be one line output. First print the minimum number (m) of cannonballs needed to wipe out the enemies followed by a single space and then m positions from which he can shoot those cannonballs. For shooting horizontally print “r” followed by the row number and for vertical shooting print “c” followed by the column number. If there is more than one solution any one will do.

 

Sample Input                               Output for Sample Input

4 4 3

1 1

1 4

3 2

 

4 4 2

1 1

2 2

 

0 0 0

 

2 r1 r3

2 r1 r2

 

---

Problemsetter: Syed Monowar Hossain

Special Thanks: Derek Kisman

 

题意：
R×C的矩阵中有N个目标，每次可消除一行或一列，求最少消除次数及其方案，多种方案输出任意一种。
分析：
显然的二分图最小覆盖，可用网络流来解。源点到每一行连一条容量为1的边，每一列到汇点连一条容量为1的边，每个目标对应行列之间的一条容量为1的边，其最小割就是方案。
最小割显然可以用最大流来解，找最后的最小割还需要在残留网络中跑一边DFS，将S部和T部分开。

#include<cstdio>#include<iostream>#include<cstdlib>#include<algorithm>#include<ctime>#include<cctype>#include<cmath>#include<string>#include<cstring>#include<stack>#include<queue>#include<list>#include<vector>#include<map>#include<set>#define sqr(x) ((x)*(x))#define LL long long#define itn int#define INF 0x3f3f3f3f#define PI 3.1415926535897932384626#define eps 1e-10#define maxm 2500007#define maxn 2007using namespace std;int fir[maxn];int u[maxm],v[maxm],cap[maxm],flow[maxm],nex[maxm];int e_max;int lv[maxn],iter[maxn],q[maxn];void add_edge(int _u,int _v,int _w){    int e;    e=e_max++;    u[e]=_u;v[e]=_v;cap[e]=_w;    nex[e]=fir[u[e]];fir[u[e]]=e;    e=e_max++;    u[e]=_v;v[e]=_u;cap[e]=0;    nex[e]=fir[u[e]];fir[u[e]]=e;}void dinic_bfs(int s){    int f,r;    memset(lv,-1,sizeof lv);    lv[s]=0;    q[f=r=0]=s;    while (f<=r)    {        int x=q[f++];        for (itn e=fir[x];~e;e=nex[e])        {            if (cap[e]>flow[e] && lv[v[e]]<0)            {                lv[v[e]]=lv[u[e]]+1;                q[++r]=v[e];            }        }    }}int dinic_dfs(int _u,int t,int _f){    if (_u==t)  return _f;    for (int &e=iter[_u];~e;e=nex[e])    {        if (cap[e]>flow[e] && lv[u[e]]<lv[v[e]])        {            int _d=dinic_dfs(v[e],t,min(_f,cap[e]-flow[e]));            if (_d>0)            {                flow[e]+=_d;                flow[e^1]-=_d;                return _d;            }        }    }    return 0;}int max_flow(int s,int t){    memset(flow,0,sizeof flow);    int total_flow=0;    for (;;)    {        dinic_bfs(s);        if (lv[t]<0)    return  total_flow;        memcpy(iter,fir,sizeof iter);        int _f;        while ((_f=dinic_dfs(s,t,INF))>0)            total_flow+=_f;    }    return total_flow;}bool vis[maxn];void dfs(int _u){    vis[_u]=true;    for (int e=fir[_u];~e;e=nex[e])    {        if (cap[e]>flow[e] && !vis[v[e]])            dfs(v[e]);    }}int main(){    #ifndef ONLINE_JUDGE        freopen("/home/fcbruce/文档/code/t","r",stdin);    #endif // ONLINE_JUDGE    int r,c,n,_u,_v;    while (scanf("%d %d %d",&r,&c,&n),r||c||n)    {        int s=0,t=r+c+1;        e_max=0;        memset(fir,-1,sizeof fir);        for (int i=1;i<=r;i++)            add_edge(s,i,1);        for (int i=r+1;i<=r+c;i++)            add_edge(i,t,1);        for (itn i=0;i<n;i++)        {            scanf("%d %d",&_u,&_v);            add_edge(_u,_v+r,1);        }        printf("%d",max_flow(s,t));        memset(vis,0,sizeof vis);        dfs(s);        for (int i=1;i<=r;i++)            if (!vis[i])    printf(" r%d",i);        for (int i=r+1;i<=r+c;i++)            if (vis[i])    printf(" c%d",i-r);        puts("");    }    return 0;}