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;
}