比赛总结，做题要细心！

今天的集训中期比赛暴露出我的很多问题，对于字符串处理的不熟悉，致使B题没写对，对于

最水的两道枚举和贪心，也是一共WA了五次，在四十几分钟之后才过的。H题的BFS思路没错，

关键代码也没写错，结果因为重复使用了变量，一直没检查出来，过了样例就胡乱提交题，

结果WA了六次，赛后才过掉。H题几乎用了三个小时，结果问题出在这个地方，心里真的不好

受！

 

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<queue>

using namespace std;



const int MAXN = 105;

const int dx[] = { 0, 0, 1, -1};

const int dy[] = { 1, -1, 0, 0};



int map[MAXN][MAXN];

char s[MAXN];

int d[MAXN][MAXN];

int n, Sx, Sy, Ex, Ey;

typedef pair<int , int> pii;



void init()

{

    int i, j;

    for( i = 1; i <= n; i ++)

    {

        scanf( "%s", s);

        for( j = 1; j <= n; j ++)

        {

            if( s[j - 1] == 'S')

            {

                map[i][j] = 0;

                Sx = i, Sy = j;

            }

            else if( s[j - 1] == 'E')

            {

                map[i][j] = 0;

                Ex = i, Ey = j;

            }

            else map[i][j] = s[j - 1] - '0';

        }

    }

}



void bfs()

{

    int x, y, nx, ny, i, j, k;

    memset( d, -1, sizeof d);

    queue<pii> q;

    pii u;

    d[Sx][Sy] = 0;

    u.first = Sx, u.second = Sy;

    q.push(u);

    while( !q.empty())

    {

        u = q.front();

        x = u.first, y = u.second;

        q.pop();

        for( k = 0; k < 4; k ++) //开始这里是i

        {

            nx = x + dx[k];

            ny = y + dy[k];

            if( d[nx][ny] < 0 && map[nx][ny] != 1 && nx >= 1 && nx <= n && ny >= 1 && ny <= n)

            {

                d[nx][ny] = d[x][y] + 1;

                if( nx == Ex && ny == Ey) return;

                u.first = nx, u.second = ny;

                q.push(u);

                if( map[nx][ny] != 0)

                {

                    for( i = 1; i <= n; i ++) //这里也是i WA了六次

                        for( j = 1; j <= n; j ++)

                        {

                            if( map[nx][ny] == map[i][j]){

                                if( d[i][j] == -1 || d[i][j] > d[nx][ny])

                                {

                                    d[i][j] = d[nx][ny];

                                    u.first = i, u.second = j;

                                    q.push(u);

                                }

                            }

                        }

                }



            }

        }

    }

}



int main()

{

    while( scanf( "%d", &n) == 1)

    {

        init();

        bfs();

        if(d[Ex][Ey] < 0)

            printf( "Oh No!\n");

        else

            printf( "%d\n", d[Ex][Ey]);

    }

    return 0;

}

 B题的字符串处理，自己写老是不对，参考标程就对了，这方面还要提高：

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>



char buf1[300], buf2[300], *a, *b;

char p[100];



int main()

{

    int i, j, t;

    while( gets(buf1) && gets(buf2))

    {

        a = buf1, b = buf2;

        sscanf( buf1, "%s", p);

        printf( "%s", p);

        for( a = strchr( a + 1, '\t'); a; a = strchr(a + 1, '\t'), ++ b)

        {

            printf("\t");

            b = strchr(b, '\t');

            if(*(a + 1) != '\t' && *(a + 1))

            {

                sscanf(a, "%s", p), printf("%s", p);

                if(*(b + 1) != '\t' && *(b + 1))

                    sscanf(b, "%s", p), printf("%s", p);

            }

        }

        printf( "\n");

    }

    return 0;

}

 E题是最短路的变种，用堆优化的Dij就能过，当时不知道怎么看成回溯的，做题还是太少了。

 

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<queue>

using namespace std;



typedef pair<int, int> pii;

const int MAXN = 5005;

const int MAXM = 20005;

const int inf = 0x3f3f3f3f;

int n, e, m;

int cost[MAXM], dist[MAXN];

int pnt[MAXM], head[MAXN], nxt[MAXM], vis[MAXN];



void addedge( int u, int v, int c)

{

    pnt[e] = v, cost[e] = c, nxt[e] = head[u], head[u] = e ++;

}



void ReadGraph()

{

    int i, a, b;

    e = 0;

    memset( head, -1, sizeof head);

    for( i = 0; i < m; i ++)

    {

        scanf( "%d%d", &a, &b);

        addedge( a, b, 0);

        addedge( b, a, 1);

    }

}



void dijkstra()

{

    priority_queue< pii, vector<pii>, greater<pii> > q;

    pii u;

    int i, x, t, y;

    memset( vis, false, sizeof vis);

    for( i = 1; i <= n; i ++)

        dist[i] = inf;

    dist[1] = 0;

    q.push( pii( 0, 1));

    while( !q.empty())

    {

        u = q.top(); q.pop();

        x = u.second, y = u.first;

        if( vis[x]) continue;

        vis[x] = true;

        for( t = head[x]; t != -1; t = nxt[t])

        {

            if( dist[pnt[t]] > y + cost[t])

            {

                dist[pnt[t]] = y + cost[t];

                q.push( pii( dist[pnt[t]], pnt[t]));

            }

        }

    }

}



int main()

{

    while( scanf( "%d%d", &n, &m) == 2)

    {

        ReadGraph();

        dijkstra();

        if( dist[n] == inf)

            printf( "-1\n");

        else

            printf( "%d\n", dist[n]);

    }

    return 0;

}