CodeForces 490E Restoring Increasing Sequence(贪心)

CodeForces 490E Restoring Increasing Sequence(贪心)

CodeForces 490E

题目大意：给N个正整数，然而这些正整数中间有些数字是被‘？’遮挡住了，每个‘？’可以还原回一个数字，希望给定的这N个整数形成一个递增的序列。可以的话，给出这N个整数的序列，不行返回N0.

解题思路：每个整数都在满足条件的情况下尽量的小，写了一个非递归版本的，发现要考虑的因素太多了，wa了太多遍。后面改成了递归版本的，但是这个比较耗时。

代码:

#include <cstdio>
#include <cstring>

const int MAXL = 10;
const int MAXN = 1e5 + 5;
char str[MAXN][MAXL];
int len[MAXN];

int N;

bool dfs (int i, int l1, int flag) {

if (l1 >= len[i]) {
if (flag)
return true;
return false;
}

if (str[i][l1] != '?') {

if (!flag && str[i][l1] < str[i - 1][l1])
return false;
return dfs(i, l1 + 1, flag | str[i][l1] > str[i - 1][l1]);
} else {

if (flag) {
str[i][l1] = '0';
if (dfs(i, l1 + 1, flag))
return true;
str[i][l1] = '?';
return false;
}

for (char k = str[i - 1][l1]; k <= '9'; k++) {
str[i][l1] = k;
if (dfs(i, l1 + 1, k > str[i - 1][l1]))
return true;
}
str[i][l1] = '?';
return false;
}
}

bool judge(int i) {

if (len[i] < len[i - 1])
return false;

if (len[i] > len[i - 1]) {
if (str[i][0] == '?')
str[i][0] = '1';
for (int j = 1; j < len[i]; j++)
if (str[i][j] == '?')
str[i][j] = '0';
return true;
}

if (dfs(i, 0, 0))
return true;
return false;
}

bool handle () {
if (N == 1)
return true;
for (int i = 1; i < N; i++)
if (!judge(i))
return false;
return true;
}

void handle_first() {
int len = strlen(str[0]);
for (int i = 0; i < len; i++)
if (str[0][i] == '?')
str[0][i] = (i == 0) ?'1' : '0';
}

int main () {

while (scanf ("%d", &N) != EOF) {

for (int i = 0; i < N; i++) {
scanf ("%s", str[i]);
len[i] = strlen(str[i]);
}

handle_first();
if (handle()) {
printf ("YES\n");
for (int i = 0; i < N; i++)
printf ("%s\n", str[i]);
} else
printf ("NO\n");
}
return 0;
}