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;
}