POJ 3114 Countries in War / HDU 2767 Proving Equivalences 强连通分量模板

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强连通分量：若在一个点集U中，对于任意两个点u,v（u,v∈U && u != v），即存在一条从 u 到 v 的路径，也存在一条从 v 到 u 的路径，则称此集合为强连通分量。

Taijan算法：

借助 index[ ] 和 rv[ ] 两个数组，rv[ u ] 记录DFS访问 u 时的时间戳，index[ u ] 记录在DFS树中，u及其子孙能通过反向边连接到的正在等待DFS回溯的点中rv[]的最小值，也可以理解成最远的祖先。

若在回溯过程中有 index[ u ] != rv[ u ] , 并将其压入栈中。

若在回溯过程中有 index[ u ] == rv[ u ],则 u 为其所在的强连通分量中第一个被访问的点。此时应将栈中的u及在u之后被压入的点弹出。

POJ 3114 ：强连通分量 + 最短路。不再赘述。

HDU 2767：若图已联通，则输出0。否则统计各连通分量形成的缩点的出入度，id，od，输出max（id,od）.

POJ 3114

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <cmath>#include <algorithm>#define LL long long#define ULL unsigned long long#define PI (acos(-1.0))#define EPS (1e-10)#pragma comment(linker,"/STACK:102400000,1024000")using namespace std;const int MAX = (1<<30);struct P{    int u,v,w;}p[250010];struct N{    int v,next;}e[250100];int index[20100];int head[20100];int rv[20100];int ty[20100];int ry[20100];int st[20100];bool id[20100];bool od[20100];int Top,Time,St_Top;void link(int u,int v){    e[Top].next = head[u];    e[Top].v = v;    head[u] = Top++;}void dfs(int s){    rv[s] = index[s] = Time++;    st[St_Top++] = s;    for(int p = head[s]; p != -1; p = e[p].next)    {        if(rv[e[p].v] == -1)        {            dfs(e[p].v);            index[s] = min(index[s],index[e[p].v]);        }        else if(ty[e[p].v] == -1)        {            index[s] = min(index[s],rv[e[p].v]);        }    }    if(index[s] == rv[s])    {        ry[Top] = s;        while(true)        {            St_Top--;            int x = st[St_Top];            ty[x] = Top;            if(x == s)                break;        }        Top++;    }}int dis[510];void sp(int u,int v,int n,int m){    if(ty[u] == ty[v])    {        printf("0\n");        return ;    }    int i,j;    bool mark;    for(i = 0;i < Top; ++i)    {        dis[i] = MAX;    }    dis[ty[u]] = 0;    for(i = 0;i < Top; ++i)    {        mark = true;        for(j = 0;j < m; ++j)        {            if(ty[p[j].v] != ty[p[j].u] && dis[ty[p[j].u]] + p[j].w < dis[ty[p[j].v]])            {                dis[ty[p[j].v]] = dis[ty[p[j].u]] + p[j].w;                mark = false;            }        }        if(mark)            break;    }    if(dis[ty[v]] == MAX)    {        printf("Nao e possivel entregar a carta\n");    }    else    {        printf("%d\n",dis[ty[v]]);    }}int main(){    int n,m,i,u,v;    while(scanf("%d %d",&n,&m) && n)    {        memset(head,-1,sizeof(int)*(n+3));        memset(rv,-1,sizeof(int)*(n+3));        memset(ty,-1,sizeof(int)*(n+3));        Top = 0;        for(i = 0; i < m; ++i)        {            scanf("%d %d %d",&p[i].u,&p[i].v,&p[i].w);            link(p[i].u,p[i].v);        }        Top = 0,Time = 0,St_Top = 0;        for(i = 1; i <= n; ++i)        {            if(rv[i] == -1)            {                dfs(i);            }        }        int k;        scanf("%d",&k);        while(k--)        {            scanf("%d %d",&u,&v);            sp(u,v,n,m);        }        printf("\n");    }    return 0;}

HDU 2767 :

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <cmath>#include <algorithm>#define LL long long#define ULL unsigned long long#define PI (acos(-1.0))#define EPS (1e-10)#pragma comment(linker,"/STACK:102400000,1024000")using namespace std;const int MAX = (1<<30);struct N{    int v,next;}e[50100];int index[20100];int head[20100];int rv[20100];int ty[20100];int ry[20100];int st[20100];bool id[20100];bool od[20100];int Top,Time,St_Top;void link(int u,int v){    e[Top].next = head[u];    e[Top].v = v;    head[u] = Top++;}void dfs(int s){    rv[s] = index[s] = Time++;    st[St_Top++] = s;    for(int p = head[s]; p != -1; p = e[p].next)    {        if(rv[e[p].v] == -1)        {            dfs(e[p].v);            index[s] = min(index[s],index[e[p].v]);        }        else if(ty[e[p].v] == -1)        {            index[s] = min(index[s],rv[e[p].v]);        }    }    if(index[s] == rv[s])    {        ry[Top] = s;        while(true)        {            St_Top--;            int x = st[St_Top];            ty[x] = Top;            if(x == s)                break;        }        Top++;    }}int main(){    //freopen("b.in","r",stdin);    int n,m,i,p,u,v,T;    scanf("%d",&T);    while(T--)    {        scanf("%d %d",&n,&m);        memset(head,-1,sizeof(int)*(n+3));        memset(rv,-1,sizeof(int)*(n+3));        memset(ty,-1,sizeof(int)*(n+3));        Top = 0;        for(i = 0; i < m; ++i)        {            scanf("%d %d",&u,&v);            link(u,v);        }        Top = 0,Time = 0,St_Top = 0;        for(i = 1; i <= n; ++i)        {            if(rv[i] == -1)            {                dfs(i);            }        }        memset(id,false,sizeof(bool)*(Top+3));        memset(od,false,sizeof(bool)*(Top+3));        for(i = 1;i <= n; ++i)        {            for(p = head[i]; p != -1; p = e[p].next)            {                if(ty[i] != ty[e[p].v])                {                    od[ty[i]] = true;                    id[ty[e[p].v]] = true;                }            }        }        int a1,a2;        for(i = 0,a1 = 0,a2 = 0;i < Top; ++i)        {            if(od[i] == false)            {                a1++;            }            if(id[i] == false)            {                a2++;            }        }        if(Top == 1)        {            printf("0\n");        }        else        {            if(a1 > a2)                printf("%d\n",a1);            else                printf("%d\n",a2);        }    }    return 0;}

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