poj1966Cable TV Network(无向图最小点割集 ISAP+邻接矩阵)
题目请戳这里
邻接表的ISAP被卡了一天。。。TLE。。。。终于被卡了。。。好忧桑啊啊啊。。。
题目大意：给一张无向图，求最少去掉几个点使图不连通。
题目分析：求无向图的点连通度，拆点建图跑最大流。具体做法是：将一个点i拆成2个点：i和i+n，分别表示从第i个点出去和进入第i个点。那么i+n->i建边，边权1，对于每一条边(a,b)，建边a->b + n,b->a+n，边权无穷。然后枚举没有边直接相连的点对(a,b)，以a为源点，b+n为汇点跑最大流，最大流量就是该图的一个割，枚举所有不相邻点对，求出最小割。具体原理就是求不相邻点对(a,b)之间的最大独立轨数目，其实就是从a出发到达b的没有公共交点的路径数目。按照上述方式建图后，每个点i和i+n之间边权为1，保证每个点只存在某一条独立轨中，这样最大流的流量就是a到b的独立轨数量。
这题一开始熟悉的邻接表+ISAP做的，怎么交都是TLE了，没办法，只好改成邻接矩阵，竟然很快过了。这题复杂度还是很高的，说明数据不强，但是邻接表为什么就TLE了呢。。。
终于发现问题了。。。原来是输入函数导致的TLE。。。
详情请见代码：
邻接表+ISAP：
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 101;
const int M = 50005;
const int inf = 0x3f3f3f3f;
int n,m,num;
struct node
{
int to,next,c,pre;
}arc[M];
int head[N],que[N],sta[N],cnt[N],dis[N],rpath[N];
int st,ed;
bool map[N][N];
int d[M][2];
void build(int s,int e,int cap)
{
arc[num].to = e;
arc[num].c = cap;
arc[num].next = head[s];
head[s] = num ++;
arc[num - 1].pre = num;
arc[num].pre = num - 1;
arc[num].to = s;
arc[num].c = 0;
arc[num].next = head[e];
head[e] = num ++;
}
void re_Bfs()
{
int i,front,rear;
for(i = 0;i < n + n;i ++)
{
dis[i] = inf;
cnt[i] = 0;
}
front = rear = 0;
cnt[0] = 1;
dis[ed] = 0;
que[rear ++] = ed;
while(front != rear)
{
int u = que[front ++];
for(i = head[u];i != -1;i = arc[i].next)
{
if(arc[arc[i].pre].c == 0 || dis[arc[i].to] < inf)
continue;
dis[arc[i].to] = dis[u] + 1;
cnt[dis[arc[i].to]] ++;
que[rear ++] = arc[i].to;
}
}
}
int ISAP()
{
re_Bfs();
int i,u,v,ret = 0;
for(i = 0;i < n + n;i ++)
sta[i] = head[i];
u = st;
while(dis[st] < n + n)
{
if(u == ed)
{
int curflow = inf;
for(i = st;i != ed;i = arc[sta[i]].to)
curflow = min(curflow,arc[sta[i]].c);
for(i = st;i != ed;i = arc[sta[i]].to)
{
arc[sta[i]].c -= curflow;
arc[arc[sta[i]].pre].c += curflow;
}
ret += curflow;
u = st;
}
for(i = sta[u];i != -1;i = arc[i].next)
if(arc[i].c > 0 && dis[u] == dis[arc[i].to] + 1)
break;
if(i != -1)
{
sta[u] = i;
rpath[arc[i].to] = arc[i].pre;
u = arc[i].to;
}
else
{
if((-- cnt[dis[u]]) == 0)
break;
sta[u] = head[u];
int tmp = n + n + 1;
for(i = sta[u];i != -1;i = arc[i].next)
if(arc[i].c > 0)
tmp = min(tmp,dis[arc[i].to]);
dis[u] = tmp + 1;
cnt[dis[u]] ++;
if(u != st)
u = arc[rpath[u]].to;
}
}
return ret;
}
int nextint()
{
int ret;
char ch;
while((ch = getchar()) > '9' || ch < '0')
;
ret = ch - '0';
while((ch - getchar()) >= '0' && ch <= '9')
ret = ret * 10 + ch - '0';
return ret;
}
void buildgraph()
{
int i,j;
memset(head,-1,sizeof(head));
num = 0;
for(i = 0;i < n;i ++)
build(i + n,i,1);
for(i = 1;i <= m;i ++)
{
build(d[i][0],d[i][1] + n,inf);
build(d[i][1],d[i][0] + n,inf);
}
}
int main()
{
int i,j;
int a,b;
//freopen("in.txt","r",stdin);
while(scanf("%d",&n) != EOF)
{
scanf("%d",&m);
memset(map,false,sizeof(map));
i = 1;
while(m --)
{
scanf(" (%d,%d)",&a,&b);
//a = nextint();b = nextint();
if(map[a][b] == false)
{
map[a][b] = map[b][a] = true;
d[i][0] = a;
d[i ++][1] = b;
}
}
m = i - 1;
int ans = inf;
for(i = 0;i < n;i ++)
{
for(j = 0;j < i;j ++)
{
if(map[i][j] == false)
{
st = i;ed = j + n;
buildgraph();
ans = min(ans,ISAP());
}
}
}
if(ans == inf)
ans = n;
printf("%d\n",ans);
}
return 0;
}
//168K 16MS 为什么那个输入函数会导致TLE呢，只是想过滤掉字符而已。下面的代码也是这样用的啊啊 啊。路过的大神求科普！！
邻接矩阵+ISAP：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 101;
const int inf = 0x3f3f3f3f;
int cap[N][N];
int flow[N][N];
bool flag[N][N];
int cnt[N],pre[N],dis[N],que[N];
int m,n,num,st,ed;
void re_Bfs()
{
int front,rear,i;
for(i = 0;i < n + n;i ++)
dis[i] = inf,cnt[i] = 0;
dis[ed] = 0;
cnt[0] = 1;
front = rear = 0;
que[rear ++] = ed;
while(front != rear)
{
int u = que[front ++];
for(i = 0;i < n + n;i ++)
if(flow[i][u] < cap[i][u] && dis[i] > n + n)
{
dis[i] = dis[u] + 1;
cnt[dis[i]] ++;
que[rear ++] = i;
}
}
}
int ISAP()
{
int i,u,ret = 0;
memset(pre,-1,sizeof(pre));
memset(flow,0,sizeof(flow));
re_Bfs();
u = st;
while(dis[st] < n + n)
{
if(u == ed)
{
int tmp = inf;
for(i = ed;i != st;i = pre[i])
tmp = min(tmp,cap[pre[i]][i] - flow[pre[i]][i]);
for(i = ed;i != st;i = pre[i])
{
flow[pre[i]][i] += tmp;
flow[i][pre[i]] -= tmp;
}
ret += tmp;
u = st;
}
for(i = 0;i < n + n;i ++)
if(cap[u][i] > flow[u][i] && dis[u] == dis[i] + 1)
break;
if(i < n + n)
{
pre[i] = u;
u = i;
}
else
{
if((-- cnt[dis[u]]) == 0)
break;
int tmp = n + n;
for(i = 0;i < n + n;i ++)
if(cap[u][i] > flow[u][i] && dis[i] + 1 < tmp)
tmp = dis[i] + 1;
dis[u] = tmp;
cnt[tmp] ++;
if(pre[u] != st)
u = pre[u];
}
}
return ret;
}
int nextint()
{
int ret;
char c;
while((c = getchar()) > '9' || c < '0')
;
ret = c - '0';
while((c = getchar()) >= '0' && c <= '9')
ret = ret * 10 + c - '0';
return ret;
}
int main()
{
int i,j;
int a,b;
while(scanf("%d%d",&n,&m) != EOF)
{
memset(cap,0,sizeof(cap));
memset(flag,false,sizeof(flag));
for(i = 0;i < n;i ++)
cap[i + n][i] = 1;
for(i = 1;i <= m;i ++)
{
a = nextint();b = nextint();
cap[a][b + n] = inf;
cap[b][a + n] = inf;
flag[a][b] = flag[b][a] = true;
}
int ans = inf;
for(i = 0;i < n;i ++)
for(j = 0;j < i;j ++)
if(flag[i][j] == false)
{
st = i;ed = j + n;
ans = min(ans,ISAP());
}
if(ans == inf)
ans = n;
printf("%d\n",ans);
}
return 0;
}
//444K 47MS