DAY 7

A，B简单

C，比较难想，但代码很简单

D，把倒着走的点提取出来，贪心，（类似差分的思路）

E,算是复习了折半搜索，但貌似爆搜加剪枝能过？？？

void sd1(int kk, long long gk) {
    if (kk >= k / 2) {
        wu = "";
        int flag = 1, ml = 0;
        for (int i = 1; i <= n; i++) {
            int hj = 0;
            for (int j = 1; j <= k / 2; j++) {
                if (a[i][j] - &#039;0&#039; == kl[j])
                    hj++;
            }
            if (hj > b[i])
                return;
            dc = (hj + &#039;0&#039;);
            wu = wu + dc;
        }
        mp[wu]++;
        lp[wu] = gk;
        return;
    }
    kk++;
    for (int i = 0; i <= 9; i++) {
        kl[kk] = i;
        sd1(kk, gk * 10 + i);
    }
    return;
}
void sd2(int kk, long long gk) {
    if (kk >= k) {
        wu = "";
        int flag = 1;
        for (int i = 1; i <= n; i++) {
            int hj = 0;
            for (int j = k / 2 + 1; j <= k; j++) {
                if (a[i][j] - &#039;0&#039; == kl[j])
                    hj++;
            }
            if (hj > b[i])
                return;
            dc = (b[i] - hj + &#039;0&#039;);
            wu = wu + dc;
        }
        if (mp[wu] == 1) {
            if (ans != -1) {
                fflag = 1;
                return;
            } else {
                ans = lp[wu] * dh[k - k / 2] + gk;
            }
        }
        if (mp[wu] > 1) {
            fflag = 1;
            return;
        }
        return;
    }
    kk++;
    for (int i = 0; i <= 9; i++) {
        kl[kk] = i;
        sd2(kk, gk * 10 + i);
        if (fflag == 1)
            return;
    }
    return;
}

F,区间最大，最小就相当于区间大于（小于）的无法放，其它区间也没法放这个值，于是就记录每个位置能放那些值，写二分图匹配

#include <bits/stdc++.h>
using namespace std;
int n, m, x, y, aa, v, op[205][205], p[205], vis[205], ans, qwq[205];
int match(int ax) {
    for (int i = 1; i <= n; i++) {
        if (op[ax][i] == 1 && vis[i] == 0) {
            vis[i] = 1;
            if (p[i] == 0 || match(p[i]) == 1) {
                p[i] = ax;
                return 1;
            }
        }
    }
    return 0;
}
int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) op[i][j] = 1;
    }
    for (int i = 1; i <= m; i++) {
        scanf("%d%d%d%d", &aa, &x, &y, &v);
        if (aa == 1) {
            for (int j = x; j <= y; j++) {
                for (int h = v + 1; h <= n; h++) {
                    op[j][h] = 0;
                }
            }
            for (int j = 1; j <= n; j++) {
                if (j < x || j > y)
                    op[j][v] = 0;
            }
        }
        if (aa == 2) {
            for (int j = x; j <= y; j++) {
                for (int h = v - 1; h >= 1; h--) {
                    op[j][h] = 0;
                }
            }
            for (int j = 1; j <= n; j++) {
                if (j < x || j > y)
                    op[j][v] = 0;
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) vis[j] = 0;
        if (match(i) == 1)
            ans++;
        else {
            cout << "-1";
            return 0;
        }
    }
    for (int i = 1; i <= n; i++) {
        qwq[p[i]] = i;
    }
    for (int i = 1; i <= n; i++) cout << qwq[i] << " ";
    return 0;
}