省选测试14

A 选择

题目大意 ： 无向图，每次删边或询问两点间是否有两条边不相交的路径

  • 没有强制在线就时光倒流一下，把删边转为加边

  • 对于判断x到y是否有两条边不相交的路径的话，可以把图建一颗生成树，然后找树边和非树边两条路径就行

  • 所以每出现一条非树边，就把两点间树边都标记一下，表示从这条边两个端点都在一个边双里，这样询问的时候就看路径中如果全被标记就存在题目要求的路径、

  • 好不容易和正解思路一模一样，但之前没写过LCT维护值的，所以Update写错了，T_T

Code

Show Code 
#include <cstdio>
#include <algorithm>
#define ls(x) t[x].c[0]
#define rs(x) t[x].c[1]
#define Get(x) (t[t[x].f].c[1] == x)
#define Nroot(x) (t[t[x].f].c[Get(x)] == x)

using namespace std;
const int N = 1e5 + 5, M = 1 << 18;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

bool ans[N];
int n, m, Q, stk[M], tp, cnt;

struct Side {
    int x, y;
}a[N];

struct Ques {
    char od; int x, y;
}q[N];

struct Hash {
    int n, x, y;
}h[M];
int head[M], hac;

void Add(int x, int y) {
    int u = (1ll * x * N + y) % M;
    h[++hac] = (Hash) {head[u], x, y}; head[u] = hac;
}

bool Find(int x, int y) {
    int u = (1ll * x * N + y) % M;
    for (int i = head[u]; i; i = h[i].n)
        if (h[i].x == x && h[i].y == y) return 1;
    return 0;
}

struct Tree {
    int f, c[2], s, sz; bool w, tr, t1;
}t[M];

void Pushup(int x) {
    t[x].sz = t[ls(x)].sz + t[rs(x)].sz + 1;
    t[x].s = t[ls(x)].s + t[rs(x)].s + t[x].w;
}

void Rotate(int x) {
    int y = t[x].f, z = t[y].f, k = Get(x), B = t[x].c[k^1];
    if (Nroot(y)) t[z].c[Get(y)] = x; t[x].f = z;
    t[x].c[k^1] = y; t[y].f = x;
    t[y].c[k] = B; t[B].f = y;
    Pushup(y); Pushup(x);
}

void Update(int x) {
    t[x].w = t[x].t1 = 1; t[x].s = t[x].sz;
}

void Pushdown(int x) {
    if (t[x].tr) {
        swap(ls(x), rs(x)); t[x].tr = 0;
        t[ls(x)].tr ^= 1; t[rs(x)].tr ^= 1;
    }
    if (t[x].t1) t[x].t1 = 0, Update(ls(x)), Update(rs(x));
}

void Splay(int x) {
    for (int p = x; p; p = Nroot(p) ? t[p].f : 0) stk[++tp] = p;
    while (tp) Pushdown(stk[tp--]);
    while (Nroot(x)) {
        int y = t[x].f;
        if (Nroot(y)) Get(x) == Get(y) ? Rotate(y) : Rotate(x);
        Rotate(x);
    }
}

void Access(int x) {
    for (int y = 0; x; y = x, x = t[x].f)
        Splay(x), t[x].c[1] = y, Pushup(x);
}

int Froot(int x) {
    Access(x); Splay(x);
    while (t[x].c[0]) x = t[x].c[0];
    return Splay(x), x;
}

void Mroot(int x) {
    Access(x); Splay(x); t[x].tr ^= 1;
}

void Split(int x, int y) {
    Mroot(x); Access(y); Splay(y);
}

void Link(int x, int y) {
    Mroot(x); t[x].f = y;
}

int main() {
    cnt = n = read(); m = read(); Q = read();
    for (int i = 1; i <= m; ++i)
        a[i] = (Side) {read(), read()};
    for (int i = 1; i <= Q; ++i) {
        char od; scanf(" %c", &od);
        int x = read(), y = read();
        q[i] = (Ques) {od, x, y};
        if (od == 'Z') Add(x, y), Add(y, x);
    }
    for (int i = 1; i <= m; ++i) {
        int x = a[i].x, y = a[i].y;
        if (Find(x, y)) continue;
        if (Froot(x) == Froot(y)) Split(x, y), Update(y);
        else ++cnt, Link(cnt, x), Link(cnt, y);
    }
    for (int i = Q; i >= 1; --i) {
        char od = q[i].od; int x = q[i].x, y = q[i].y;
        if (od == 'P') ans[i] = (Froot(x) == Froot(y) && (Split(x, y), t[y].sz == t[y].s));
        else if (Froot(x) == Froot(y)) Split(x, y), Update(y);
        else ++cnt, Link(cnt, x), Link(cnt, y);
    }
    for (int i = 1; i <= Q; ++i)
        if (q[i].od == 'P') puts(ans[i] ? "Yes" : "No");
    return 0;
}

B 三元组

题目大意 ： 所有相邻的回文串最左端和最右端乘积的和

  • 很容易想到枚举两个回文串的间隙(x,x+1)在哪，然后求出以x为结尾的回文串的左端点的和跟以x+1为开头的回文串右端点的和，然后乘起来，但是这两个东西不好求

  • i可以拆成x-len0+1, k可以拆成x+len1，要求出所有的i的和跟k的和，只要求出以一个点为结尾的字符串的个数和长度和就行，用回文自动机可以做到

Code

Show Code 
#include <cstdio>
#include <cstring>
#include <algorithm>
#define int long long
using namespace std;
const int N = 1e6 + 5, M = 1e9 + 7;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >='0' && c <='9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

char c[N];
int n, fail[N], t[N][26], len[N], s[N], dep[N], sl[N], sr[N], cl[N], cr[N], trc;

int Find(int x, int y) {
    while (c[y-len[x]-1] != c[y]) x = fail[x];
    return x;
}

void Pam(int *S, int *C) {
    memset(t, 0, sizeof(t[0]) * (trc + 1)); 
    int x = trc = 1;
    for (int i = 1; i <= n; ++i) {
        int y = c[i] - 'a'; x = Find(x, i);
        if (!t[x][y]) {
            ++trc;
            len[trc] = len[x] + 2;
            fail[trc] = t[Find(fail[x], i)][y];
            s[trc] = (s[fail[trc]] + len[trc]) % M;
            dep[trc] = dep[fail[trc]] + 1;
            t[x][y] = trc;
        }
        x = t[x][y];
        S[i] = s[x]; C[i] = dep[x];
    }
}

signed main() {
    int T = read();
    fail[0] = 1; c[0] = len[1] = -1;
    while (T--) {
        scanf("%s", c + 1);
        n = strlen(c + 1);
        Pam(sr, cr);
        reverse(c + 1, c + n + 1);
        Pam(sl, cl);
        reverse(sl + 1, sl + n + 1);
        reverse(cl + 1, cl + n + 1);
        int ans = 0;
        for (int i = 1; i < n; ++i) {
            int r = (1ll * cr[i] * (i + 1) % M - sr[i] + M) % M;
            //cr[i] 是以 i 为结尾的回文串个数，sr[i] 是 以 i 为结尾的回文串长度和
            int l = (1ll * cl[i+1] * i % M + sl[i+1]) % M;
            //cl[i] 是以 i 为开头的回文串个数，sl[i] 是 以 i 为开头的回文串长度和
            if ((ans += 1ll * r * l % M) >= M) ans -= M;
        }
        printf("%lld
", ans);
    }
}

C 最有价值

题目大意 ： 对于题中定义的价值，让选出一个子序列使得价值最大

  • 最大权闭合图

  • 对于 (w_{i,j}+w+{j,i}) 建一个点，对于每个位置建一个点，代价为 -a,对于每个字符建一个点，代价为 -b+a,

  • 一类点连向二类点，二类点连向三类点

Code

Show Code 
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 1e4 + 5;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar())
        if (c == '-') f = -1;
    for (; c >='0' && c <='9'; c = getchar())
        x = x * 10 + c - '0';
    return x * f;
}

struct Edge {
    int n, t, d;
}e[N*4];
int h[N], edc;

void Add(int x, int y, int d) {
    e[++edc] = (Edge) {h[x], y, d}; h[x] = edc;
    e[++edc] = (Edge) {h[y], x, 0}; h[y] = edc;
}

char c[N];
int n, a[15], b[15], w[105][105], ans;
int s, t, cnt, dep[N], q[N], l, r, tmp[N];

bool Bfs() {
    memcpy(h + 1, tmp + 1, cnt * 4);
    memset(dep + 1, 0, cnt * 4);
    dep[s] = 1; q[l=r=1] = s;
    while (l <= r) {
        int x = q[l++];
        for (int i = h[x], y; i; i = e[i].n) {
            if (!e[i].d || dep[y=e[i].t]) continue;
            dep[y] = dep[x] + 1; q[++r] = y;
            if (y == t) return 1;
        }
    }
    return 0;
}

int Dinic(int x, int lim) {
    if (x == t) return lim;
    int sum = 0;
    for (int i = h[x]; i && lim; i = e[i].n) {
        int y = e[i].t; h[x] = i;
        if (!e[i].d || dep[y] != dep[x] + 1) continue;
        int f = Dinic(y, min(lim, e[i].d));
        sum += f; lim -= f;
        e[i].d -= f; e[i^1].d += f;
    }
    if (!sum) dep[x] = 0;
    return sum;
}

int main() {
    int T = read();
    while (T--) {
        memset(h + 1, 0, cnt * 4); edc = 1;
        n = read(); ans = 0; cnt = n + 15;
        s = ++cnt; t = ++cnt;
        scanf("%s", c + 1);
        for (int i = 0; i <= 9; ++i) {
            a[i] = read(), b[i] = read(); 
            Add(s, i + 1 + n, b[i] - a[i]);
        }
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                w[i][j] = read();
        for (int i = 1; i <= n; ++i) {
            Add(s, i, a[c[i]-'0']);
            Add(c[i] - '0' + 1 + n, i, 1e9);
            for (int j = i + 1; j <= n; ++j) {
                cnt++; ans += w[i][j] + w[j][i];
                Add(i, cnt, 1e9); Add(j, cnt, 1e9);
                Add(cnt, t, w[i][j] + w[j][i]);
            }
        }
        memcpy(tmp + 1, h + 1, cnt * 4);
        while (Bfs()) ans -= Dinic(s, 1e9);
        printf("%d
", ans);
    }
    return 0;
}
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原文地址：https://www.cnblogs.com/shawk/p/14410466.html