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  • 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
    ",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
    ",ans);
        }
        return 0;
    }
    //444K	47MS


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  • 原文地址:https://www.cnblogs.com/keanuyaoo/p/3331409.html
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