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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


 

 

posted on 2013-09-21 13:05  风言枫语  阅读(250)  评论(0编辑  收藏  举报