POJ 2570 Fiber Network (传递闭包)

题意：给定n个城市之间的有向图，每条边是以字母标示的，输出在每条可行线路u->v中，其所经过的每一段线路的相同的字母。

思路：由于两点之间有多条线路，所以想到了对每个字母进行floyd，最后集中输出，然后华丽的TLE了.......后来知道可以用位运算优化，汗......没想出来.....然后就可以AC了。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
using namespace std;

const int BORDER = (1<<20)-1;
const int MAXSIZE = 37;
const int MAXN = 505;
const int INF = 1000000000;
#define CLR(x,y) memset(x,y,sizeof(x))
#define ADD(x) x=((x+1)&BORDER)
#define IN(x) scanf("%d",&x)
#define OUT(x) printf("%d\n",x)
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) ((x)>0?(x):-(x))


int map[MAXN][MAXN];
int n,m;
bool mark[MAXN];

int init()
{
    CLR(map,0);
    CLR(mark,0);
    return 0;
}
int input()
{
    int i,u,v,k,len;
    char str[30];
    for( ; ; )
    {
        scanf("%d%d",&u,&v);
        if(!u || !v)
            break;
        scanf("%s",str);
        len = strlen(str);
        for(i = 0; i < len; ++i)
            map[u][v] |= (1<<(str[i]-'a'));
    }
    return 0;
}
int floyd()
{
    int i,j,k,tmp;
    for(k = 1; k <= n; ++k)
        for(i = 1; i <= n; ++i)
            for(j = 1; j <= n; ++j)
                map[i][j] |= (map[i][k] & map[k][j]);
    return 0;
}
int work()
{
    int i,j,tmp,u,v;
    bool tag;
    floyd();
    for( ; ; )
    {
        scanf("%d%d",&u,&v);
        if(!u || !v)
            break;
        tag =false;
        for(i = 0; i < 26; ++i)
            if(map[u][v] & (1<<i))
            {
                tag = true;
                printf("%c",'a'+i);
            }
        if(tag)
            printf("\n");
        else
            printf("-\n");
    }
    printf("\n");
    return 0;
}
int main()
{
    while(IN(n))
    {
        if(!n)
            break;
        init();
        input();
        work();
    }
    return 0;

TLE code：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
using namespace std;

const int BORDER = (1<<20)-1;
const int MAXSIZE = 37;
const int MAXN = 505;
const int INF = 1000000000;
#define CLR(x,y) memset(x,y,sizeof(x))
#define ADD(x) x=((x+1)&BORDER)
#define IN(x) scanf("%d",&x)
#define OUT(x) printf("%d\n",x)
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) ((x)>0?(x):-(x))

typedef struct{
    bool alp[26];
}Node;

Node map[MAXN][MAXN];
int n,m;
bool mark[MAXN];

int init()
{
    CLR(map,0);
    CLR(mark,0);
    return 0;
}
int input()
{
    int i,u,v,k,len;
    char str[30];
    for( ; ; )
    {
        scanf("%d%d",&u,&v);
        if(!u || !v)
            break;
        scanf("%s",str);
        len = strlen(str);
        for(i = 0; i < len; ++i)
        {
            map[u][v].alp[str[i]-'a'] = true;
            mark[str[i]-'a'] = true;
        }
    }
    return 0;
}
int floyd(const int& ind)
{
    int i,j,k,tmp;
    for(k = 1; k <= n; ++k)
        for(i = 1; i <= n; ++i)
            for(j = 1; j <= n; ++j)
                map[i][j].alp[ind] |= (map[i][k].alp[ind]&map[k][j].alp[ind]);
    return 0;
}
int work()
{
    int i,j,tmp,u,v;
    bool tag;
    for(i  = 0; i < 26; ++i)
    {
        if(!mark[i])
            continue;
        floyd(i);
    }
    for( ; ; )
    {
        scanf("%d%d",&u,&v);
        if(!u || !v)
            break;
        tag =false;
        for(i = 0; i < 26; ++i)
            if(map[u][v].alp[i])
            {
                tag = true;
                printf("%c",'a'+i);
            }
        if(tag)
            printf("\n");
        else
            printf("-\n");
    }
    return 0;
}
int main()
{
    while(IN(n))
    {
        if(!n)
            break;
        init();
        input();
        work();
    }
    return 0;
}