NOIP模拟测试8

HZ怎么老考试啊23333

考试前一天占个坑。给自己的忠告：想不出正解就别想了，暴力打满Rank就不会难看QaQ

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考完了，Rank14，我又boomboomboom了orz

最后T3的暴力还是没打出来，我就是个想不出正解还要ning想的DD

T1:匹配

我是Sb

一眼KMP，刚正不阿的cwy不会忘记了看猫片，果断\(O(N^2)\)暴力走人（

后来一看正解居然是蛤希，我草了，，，早知道就ning干力QAQ

考场上现推KMP估计还是能推出来的，如果能多想想就好了

简直对不起图巨

而且想不到KMP为啥不想想别的啊。。。懊悔中

复习下KMP：

\(nxt[i]\)表示了\(S[,i]\)的前缀最长border的位置，当我们要匹配一个新的字母，我们设定一个指针p = nxt[i - 1]，如果S[i] = S[p+1]说明匹配大成功，p+1就是\(S[,i]\)的前缀最长border的位置，否则p指针一直往回跳，最恶情况就是无法匹配，一直跳到了1。

#include <bits/stdc++.h>

const int N = 300000 + 233;
int T, la, lb, nxt[N];
char a[N], b[N], add[5];

signed main() {
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &la, &lb);
        scanf("%s%s", a + 1, add);
        a[la + 1] = '$', ++la;
        for (int i = 1; i <= lb; i++) a[i + la] = a[i];
        a[la + lb + 1] = add[0];
        for (int i = 2; a[i]; i++) {
            int p = nxt[i - 1];
            while (p && a[i] != a[p + 1]) p = nxt[p];
            if (a[i] == a[p + 1]) nxt[i] = p + 1;
        }
        printf("%d\n", nxt[la + lb + 1]);
        memset(nxt, 0, sizeof(nxt));
    }
    return 0;
}

T2:回家

我离AC只差一个90w的数组

一眼求割点，二眼好像不对，割点不一定在1-n路径上。于是建个圆方树，dfs求出1-n经过的圆点。

然后就MLE了？？？？红太阳GMK说他数组开小了，我跟风给常数乘个10.

然后我就A了？？？

这题还是挺显然的，思路秒出

考试时候千万别吝啬空间，够开就开，QAQ

#include <bits/stdc++.h>

const int N = 2000000 + 233;
int T, n, m, ans;
struct Edge {int to, nxt;} e[N << 2], c[N << 2];
int ecnt, head[N], ccnt, hc[N], tot;
bool yes[N];

inline int R() {
    int a = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) a = a * 10 + c - '0', c = getchar();
    return a;
}

int low[N], dfn[N], num, stk[N], p;

inline void AddEdge(int f, int to) {
    e[++ecnt] = {to, head[f]}, head[f] = ecnt;
}

inline void AddC(int f, int to) {
    c[++ccnt] = {to, hc[f]}, hc[f] = ccnt;
}

void Tarjan(int x) {
    dfn[x] = low[x] = ++num;
    stk[++p] = x;
    for (int i = head[x], y = e[i].to; i; i = e[i].nxt, y = e[i].to) {
        if (!dfn[y]) {
            Tarjan(y);
            low[x] = std::min(low[x], low[y]);
            if (dfn[x] == low[y]) {
                AddC(x, ++tot), AddC(tot, x); int z;
                do {
                    z = stk[p--], AddC(z, tot), AddC(tot, z);
                } while (y != z);
            }
        } else low[x] = std::min(low[x], dfn[y]);
    }
}

void dfs(int x, int fa) {
    for (int i = hc[x], y = c[i].to; i; i = c[i].nxt, y = c[i].to) {
        if (y != fa) {
            dfs(y, x);
            if (y == n) return (void) (yes[x] = 1);
            else if (yes[y]) yes[x] = 1;
        }
    }
}

signed main() {
    T = R();
    while (T--) {
        n = R(), m = R(), tot = n;
        for (int i = 1, x, y; i <= m; i++)
            x = R(), y = R(), AddEdge(x, y), AddEdge(y, x);
        Tarjan(1);
        dfs(1, 0);
        for (int i = 2; i < n; i++)
            if (yes[i]) ans++;
        printf("%d\n", ans);
        for (int i = 2; i < n; i++)
            if (yes[i]) printf("%d ", i);
        if (ans) printf("\n");
        memset(head, 0, sizeof(head));
        memset(low, 0, sizeof(low));
        memset(dfn, 0, sizeof(dfn));
        memset(hc, 0, sizeof(hc));
        memset(stk, 0, sizeof(stk));
        memset(yes, 0, sizeof(yes));
        num = ecnt = ccnt = ans = p = 0;
    }
    return 0;
}