分享一下华子8月31号机考代码
仅供参考
第一题基本的字符串操作，用二维数组存放句子和单词表里的单词，注意大小写不敏感，最后遍历搜索即可；
第二题深度优先搜索算法，注意回溯的时候炸弹炸掉的墙要复原。且炸弹和陷阱只可以使用一次。
第三题动态规划，注意dp数组的含义，我设的dp[i][j]数组包含了在第i个充电桩充电和等待的时间，以及到达下一个充电桩的路途中所使用的时间。然后倒叙遍历。正序遍历应可以，不过dp数组的含义会不一样。注意边界初始化条件即可。
第一题：
#include<stdio.h>
#include<string.h>
int main(void)
{
char str[10005]={0};
char sentence[250][60]={0};
char backupsentence[250][60]={0};
scanf("%[^\n]", &str);
getchar();
char word[2100][60] = { 0 };
char str2[100001] = { 0 };
scanf("%[^\n]", &str2);
int len1 = strlen(str);
int j1 = 0;
int j2 = 0;
int jj1 = 0;
int jj2 = 0;
for (int i = 0; i < len1; i++)
{
if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))
{
while (!(str[i] == ' ' || str[i] == ',' || str[i] == '.') || str[i] == '"')
{
sentence[j1][j2++] = str[i];
backupsentence[jj1][jj2++] = str[i];
i++;
}
i--;
for (int m = 0; m < j2; m++)
{
if (sentence[j1][m] >= 'A' && sentence[j1][m] <= 'Z') { sentence[j1][m] += 32; }
}
j1++; j2 = 0;
jj1++; jj2 = 0;
}
if (str[i] == ' '|| str[i] == ',' || str[i] == '.')
{
sentence[j1][j2++] = str[i];
backupsentence[jj1][jj2++] = str[i];
j1++; j2 = 0;
jj1++; jj2 = 0;
}
if (str[i] == '"')
{
sentence[j1][j2++] = str[i];
backupsentence[jj1][jj2++] = str[i];
i++;
while (str[i] != '"')
{
sentence[j1][j2++] = str[i];
backupsentence[jj1][jj2++] = str[i];
i++;
}
for (int m = 1; m < j2; m++)
{
if (sentence[j1][m] >= 'A' && sentence[j1][m] <= 'Z') { sentence[j1][m] += 32; }
}
sentence[j1][j2++] = str[i];
backupsentence[jj1][jj2++] = str[i];
j1++; j2 = 0;
jj1++; jj2 = 0;
}
}//sentence输入完毕
int list = j1;//包含标点符号，sentence中单词的数量
/*for (int i = 0; i < list; i++)
{
printf("%s\n",backupsentence[i]);
}*/
int len2 = strlen(str2);
int k1 = 0;
int k2 = 0;
for (int i = 0; i < len2; i++)
{
if ((str2[i] >= 'a' && str2[i] <= 'z') || (str2[i] >= 'A' && str2[i] <= 'Z'))
{
while (str2[i] != ' ')
{
word[k1][k2++] = str2[i];
i++;
}
i--;
for (int m = 0; m < k2; m++)
{
if (word[k1][m] >= 'A' && word[k1][m] <= 'Z') { word[k1][m] += 32; }
}
k1++; k2 = 0;
}
if (str2[i] == ' ')
{
//word[k1][k2++] = str2[i];
//k1++; k2 = 0;
}
}
int index = k1;
/*for (int i = 0; i < index; i++)
{
printf("%s\n", word[i]);
}*/
//printf("%d", list);
//printf("%d\n", index);
int replace[250] = { 0 };
int num[250] = { 0 };
for (int i = 0; i < list; i++)
{
int maxindex = 0;
if (strcmp(sentence[i], " ") != 0 && strcmp(sentence[i], ",") != 0 && strcmp(sentence[i], ".") != 0)
{
for (int m = 0; m < index; m++)
{
if (strcmp(sentence[i], word[m]) == 0)
{
replace[i] = 1;
maxindex = (maxindex > m) ? maxindex : m;
}
num[i] = maxindex;
}
}
}
for (int i = 0; i < list; i++)
{
if (replace[i] == 0)
{
printf("%s", backupsentence[i]);
}
else if (replace[i] == 1)
{
printf("%d", num[i]);
}
}
return 0;
} 第二题： #include<stdio.h>
#include<string.h>
int n, m;
int map[30][30] = { 0 };
int startx, starty;
int endx, endy;
int mintime = 999;
//int time=0;
int vis[30][30] = { 0 };
int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };
void dfs(int x, int y, int time)
{
int tx, ty;
if (x == endx && y == endy)
{
mintime = (mintime > time) ? time : mintime;
return;
}
for (int k = 0; k < 4; k++)
{
tx = x + dx[k];
ty = y + dy[k];
if (tx >= 0 && tx < n && ty >= 0 && ty < m && vis[tx][ty] == 0)
{
if (map[tx][ty] == 0||map[tx][ty]==3||map[tx][ty]==2)
{
vis[tx][ty] = 1;
dfs(tx, ty, time + 1);
vis[tx][ty] = 0;
}
else if (map[tx][ty] == 4)
{
map[tx][ty] = 0;//陷阱只可以发挥一次作用
vis[tx][ty] = 1;
dfs(tx, ty, time + 3);
vis[tx][ty] = 0;
map[tx][ty] = 4;//陷阱复原
}
else if (map[tx][ty] == 6)
{
map[tx][ty] = 0;//炸弹只可以发挥一次作用
vis[tx][ty] = 1;
int memx[10] = { 0 };
int memy[10] = { 0 };
int j1 = 0;
for (int m = 0; m < 4; m++)
{
if (map[tx + dx[m]][ty + dy[m]] == 1)
{
memx[j1++] = tx + dx[m];
memy[j1++] = ty + dy[m];
map[tx + dx[m]][ty + dy[m]] = 0;
}
}
dfs(tx, ty, time + 1);
map[tx][ty] = 6;//炸弹复原
for (int m = 0; m < j1; m++)
{
map[memx[m]][memy[m]] = 1;//炸毁的墙复原
}
}
}
}
return;
}
int main(void)
{
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
scanf("%d", &map[i][j]);
if (map[i][j] == 2) { startx = i; starty = j; }
if (map[i][j] == 3) { endx = i; endy = j; }
}
}//输入完毕
/*startx = 0;
starty = 0;
endx = n - 1;
endy = m - 1;*/
vis[startx][starty] = 1;
dfs(startx, starty, 0);
//printf("%d %d\n", startx,starty);
//printf("%d %d\n", endx, endy);
printf("%d", mintime);
return 0;
} 第三题：
#include<stdio.h>
#include<string.h>
//二维dp数组解法
int D;
int N;
typedef struct {
int distance;
int time;
}STOP;
int dp[10005][1010] = { 0 };
int dis = 0;
int main(void)
{
scanf("%d%d", &D, &N);
STOP stop[10005]={0};
for (int i = 1; i<=N; i++)
{
scanf("%d%d", &stop[i].distance, &stop[i].time);
}//输入完毕
stop[N + 1].distance = D;
stop[N + 1].time = 0;
//printf("%d %d", stop[0].distance, stop[0].time);
//两个特殊情况，限制两个站点之间距离小于1000
for (int i = 0; i < N; i++)
{
if (stop[i + 1].distance - stop[i].distance > 1000) { printf("-1"); return 0; }
}
if (D - stop[N].distance > 1000) { printf("-1"); return 0; }
//int dp[i][j]表示在到达第i个站点，剩余续航公里数j，到达下一个站点前花掉的时间
/*if(stop[i+1].distance-stop[i].distance>j),dis=stop[i+1].distance-stop[i].distance;
* 此时在第i个站点必须充电，dp[i][j]=dp[i+1][1000-dis]+dis/100+stop[i].time+1;
* if(dis<=j)可以选择充电或者不充，如果充电则dp[i][j]=dp[i+1][1000-dis]+dis/100+stop[i].time+1;
* 如果选择不充电，则dp[i][j]=dp[i+1][j-dis]+dis/100;
//*/
//初始化dp
for (int i = 0; i <= 1000; i += 100)
{
int temp = 0;temp = D - stop[N].distance;
if(i>temp){ dp[N][i] = temp / 100; }
else
{
dp[N][i] = temp / 100 + stop[N].time + 1;
}
}
//动态规划
for (int i = N-1; i > 0; i--)
{
dis = stop[i + 1].distance - stop[i].distance;
for (int j = 0; j <= 1000; j += 100)
{
if(j<dis)
{
dp[i][j] = dp[i + 1][1000 - dis] + dis / 100 + stop[i].time + 1;
}
if (j >= dis)
{
int temp1; temp1= dp[i + 1][1000 - dis] + dis / 100 + stop[i].time + 1;
int temp2; temp2 = dp[i + 1][j - dis] + dis / 100;
dp[i][j] = (temp1 > temp2) ? temp2 : temp1;
}
}
}
printf("%d",dp[1][1000-stop[1].distance]+stop[1].distance/100);
return 0;
} 
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