Friendship
Description
In
modern society, each person has his own friends. Since all the people
are very busy, they communicate with each other only by phone. You can
assume that people A can keep in touch with people B, only if
1. A knows B's phone number, or
2. A knows people C's phone number and C can keep in touch with B.
It's assured that if people A knows people B's number, B will also know A's number.
Sometimes, someone may meet something bad which makes him lose
touch with all the others. For example, he may lose his phone number
book and change his phone number at the same time.
In this problem, you will know the relations between every two
among N people. To make it easy, we number these N people by 1,2,...,N.
Given two special people with the number S and T, when some people meet
bad things, S may lose touch with T. Your job is to compute the minimal
number of people that can make this situation happen. It is supposed
that bad thing will never happen on S or T.
Input
The
first line of the input contains three integers N (2<=N<=200), S
and T ( 1 <= S, T <= N , and S is not equal to T).Each of the
following N lines contains N integers. If i knows j's number, then the
j-th number in the (i+1)-th line will be 1, otherwise the number will
be 0.
You can assume that the number of 1s will not exceed 5000 in the input.
Output
If
there is no way to make A lose touch with B, print "NO ANSWER!" in a
single line. Otherwise, the first line contains a single number t,
which is the minimal number you have got, and if t is not zero, the
second line is needed, which contains t integers in ascending order
that indicate the number of people who meet bad things. The integers
are separated by a single space.
If there is more than one solution, we give every solution a score,
and output the solution with the minimal score. We can compute the
score of a solution in the following way: assume a solution is A1, A2,
..., At (1 <= A1 < A2 <...< At <=N ), the score will be
(A1-1)*N^t+(A2-1)*N^(t-1)+...+(At-1)*N. The input will assure that
there won't be two solutions with the minimal score.
Sample Input
3 1 3
1 1 0
1 1 1
0 1 1
Sample Output
1
2
题意：给出一些人之间的通话关系，如果A能跟B联系，B能跟C联系，那么A能跟C联系，但如果这条路径上A跟B断了联系，除非A通过其他人能联系上C，否则就联系不上C了。
求删除最少的人，然给定的两个点断了联系，如果存在被删除的店，要求输出字典序最小的。就是编号小的优先。分析：题目是求图上s到t的连通度，即从s到t存在多少条独立轨（即点不相交的路径，点不相交，则边当然不相交了，点在边才在吗）。这些路径上每条都删掉一个点（非s非t的点），就可以让s跟t失去了关系。由于是独立轨，所以每个点只能经过一个流量。每个x点拆成两个点x，x'。流量为1。原图中的边流量无穷。题目要求小点优先，所以从小到大枚举每个点被删除时产生的最大流是否比之前的小。如果是，则这个点肯定得被删除。代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAXN = 1005;
const int MAXM = 210000;
const int INF = 1000000000;
struct Edge
{
int st, ed;
int next;
int flow;
int cap;
}edge[MAXM];
int head[MAXN], d[MAXN], hash[MAXN], map[300][300];
int value[MAXN];
int N, M, F, E;
void add(int u, int v, int w)
{
edge[E].flow = 0;
edge[E].cap = w;
edge[E].st = u;
edge[E].ed = v;
edge[E].next = head[u];
head[u] = E++;
edge[E].flow = 0;
edge[E].cap = 0;
edge[E].st = v;
edge[E].ed = u;
edge[E].next = head[v];
head[v] = E++;
}
int dinic_bfs(int src, int dest, int ver)
{
int i, j;
for (i = 0; i <= ver; i++)
{
if (hash[i])//标记不要点，设坏它们的level值
{
d[i] = -2;
}
else d[i] = -1;
}
int que[MAXN], rear = 1;
que[0] = src; d[src] = 0;
for(i = 0; i < rear; i++)//队列
{
for(j = head[que[i]]; j != -1; j = edge[j].next)
{
if(d[edge[j].ed] == -1 && edge[j].cap > edge[j].flow)
{
d[edge[j].ed] = d[que[i]]+1;
que[rear++] = edge[j].ed;
}
}
}
return d[dest] >= 0;
}
int dinic_dfs(int src, int dest, int ver)
{
int stk[MAXN], top = 0;
int ret = 0, cur, ptr, pre[MAXN], minf, i;
int del[MAXN], out[MAXN];
for (i = 0; i <= ver; i++)
{
del[i] = 0, out[i] = head[i];
}
stk[top++] = src;
pre[src] = src;
cur = src;
while(top)
{
while(cur != dest && top)
{
for(i = out[cur]; i != -1; i = edge[i].next)
{
if(d[edge[i].ed] == d[cur] + 1 && edge[i].cap > edge[i].flow && !del[edge[i].ed])
{
stk[top++] = edge[i].ed;
cur = edge[i].ed;
pre[edge[i].ed] = i;
break;
}
}
if(i == -1)//该节点的所有邻接点都被访问，则将该节点
{
del[cur] = 1;
top--;
if(top) cur = stk[top-1];
}
}
if(cur == dest)
{
minf = INF;
while(cur != src)
{
cur = pre[cur];
if(edge[cur].cap - edge[cur].flow < minf) minf = edge[cur].cap - edge[cur].flow;
cur = edge[cur].st;
}
cur = dest;
while(cur != src)
{
cur = pre[cur];
edge[cur].flow += minf;
edge[cur^1].flow -= minf;
if(edge[cur].cap - edge[cur].flow == 0)
{
ptr = edge[cur].st;
}
cur = edge[cur].st;
}
while(top > 0&& stk[top-1] != ptr) top--;
if(top) cur = stk[top-1];
ret += minf;
}
}
return ret;
}
int Dinic(int src, int dest, int ver)
{
int ret = 0, t;
while(dinic_bfs(src, dest, ver))
{
t = dinic_dfs(src, dest, ver);
if(t) ret += t;
else break;
}
return ret;
}
int main()
{
int n, s, t, i, j, w, ans, m;
int src, dest, ver;
while (scanf("%d%d%d", &n, &s, &t) - EOF)
{
m = n + n;
ver = n + n;//顶点总数
for (i = 0, E = 0; i <= m; i++)
{
head[i] = -1;
hash[i] = 0;
}
for (i = 1; i <= n; i++)
{
if (s == i)
{
add(s, s + n, INF);
}
else if (i == t) add(t, t + n, INF);
else
{
add(i, i + n, 1), add(i + n, i, 1);
}
}
for (i = 1; i <= n; i++)
{
for (j = 1; j <= n; j++)
{
scanf("%d", &map[i][j]);
if (map[i][j] && i != j && i != t && j != s)
{
add(i + n, j, INF);
}
}
}
if (map[s][t])
{
printf("NO ANSWER!\n");
continue;
}
src = s, dest = t + n;//源跟汇
ans = Dinic(src, dest, ver);
printf("%d\n", ans);
if (ans == 0) continue;
int pre, x, k;
for (k = 1, pre = ans; k <= n; k++)
{
if (k == s || k == t) continue;
for (i = 0; i < E; i++)
{
edge[i].flow = 0;
}
hash[k] = 1;
x = Dinic(src, dest, ver);
if (x < pre)
{
pre = x;
}
else hash[k] = 0;
}
int sign = 0;
for (i = 1; i <= n; i++)
{
if (hash[i])
{
if (sign) printf(" ");
printf("%d", i);
sign = 1;
}
}
if (sign) printf("\n");
}
return 0;
}
/*
2 1 2
1 1
1 1
10 10 1
0 1 1 1 1 0 1 1 0 0
1 0 0 0 0 1 1 1 1 1
1 0 0 1 1 1 1 1 1 1
1 0 1 0 1 1 0 1 1 0
1 0 1 1 0 0 1 1 1 1
0 1 1 1 0 0 1 0 0 1
1 1 1 0 1 1 0 1 1 0
1 1 1 1 1 0 1 0 0 0
0 1 1 1 1 0 1 0 0 1
0 1 1 0 1 1 0 0 1 0
10 1 10
0 1 1 1 1 0 1 1 0 0
1 0 0 0 0 1 1 1 1 1
1 0 0 1 1 1 1 1 1 1
1 0 1 0 1 1 0 1 1 0
1 0 1 1 0 0 1 1 1 1
0 1 1 1 0 0 1 0 0 1
1 1 1 0 1 1 0 1 1 0
1 1 1 1 1 0 1 0 0 0
0 1 1 1 1 0 1 0 0 1
0 1 1 0 1 1 0 0 1 0
4 1 4
1 1 1 0
1 1 0 1
1 0 1 1
0 1 1 1
4 4 1
1 1 1 0
1 1 0 1
1 0 1 1
0 1 1 1
*/