Hdu5068线段树

题意 ：每层都有2个门，每个门后面都能通向下一层的两个门，有两个操作 ，第一个 问 从a层到 b层的方法数，第二个 修改 其中某门通向某门的状态。

 

线段树单点更新，矩阵连通性构造 离散数学里有讲。

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <climits>

#include <string>

#include <iostream>

#include <map>

#include <cstdlib>

#include <list>

#include <set>

#include <queue>

#include <stack>

#include <math.h>

using namespace std;

#define lson l,mid,rt<<1

#define rson mid+1,r,rt<<1|1

typedef long long LL;

const LL maxn = 55555;

const LL mod = 1000000007;

struct Node

{

    LL x1, x2, x3, x4;

};

Node ans;

Node node[maxn << 2];

Node gao(Node a, Node b)

{

    Node t;

    t.x1 = a.x1*b.x1%mod + a.x2*b.x3%mod;

    t.x2 = a.x1*b.x2%mod + a.x2*b.x4%mod;

    t.x3 = a.x3*b.x1%mod + a.x4*b.x3%mod;

    t.x4 = a.x3*b.x2%mod + a.x4*b.x4%mod;

    return t;

}



void up(LL rt)

{

    node[rt] = gao(node[rt<<1],node[rt<<1|1]);

}



void build(LL l, LL r, LL rt)

{

    if (l == r){

        node[rt].x1 = 1; node[rt].x2 = 1; node[rt].x3 = 1; node[rt].x4 = 1;

        return;

    }

    LL mid = (l + r) >> 1;

    build(lson);

    build(rson);

    up(rt);

}

void init()

{

    ans.x1 = 1; ans.x2 = 0; ans.x3 = 0; ans.x4 = 1;

}

void update(LL k, LL i, LL l, LL r, LL rt)

{

    if (l == r){

        switch (i){

        case 1:node[rt].x1 ^= 1; break;

        case 2:node[rt].x2 ^= 1; break;

        case 3:node[rt].x3 ^= 1; break;

            default:node[rt].x4 ^= 1;

        }

        return;

    }

    LL mid = (l + r) >> 1;

    if (k <= mid) update(k, i, lson);

    if (k > mid) update(k, i, rson);

    up(rt);

}



void  ask(LL L, LL R, LL l, LL r, LL rt)

{

    if (L <= l&&r <= R){

        ans = gao(ans, node[rt]);

        return;

    }

    LL mid = (l + r) >> 1;

    if (L <= mid) ask(L, R, lson);

    if (R > mid) ask(L, R, rson);



}



void output()

{

    LL sum;

    sum = (ans.x1 + ans.x2)%mod + (ans.x3 + ans.x4)%mod;

    sum%=mod;

    printf("%I64d\n", sum);

}

int main()

{

    LL a, b, c, t, n, m;

    while (cin >> n >> m){

        build(1, n - 1, 1);

        for (LL i = 0; i < m; i++){

            scanf("%I64d", &t);

            if (t == 0){

                scanf("%I64d%I64d", &a, &b);

                if (a == b){

                    printf("0\n"); continue;

                }

                init();

                ask(a, b - 1, 1, n - 1, 1);

                output();

            }

            else{

                scanf("%I64d%I64d%I64d", &a, &b, &c);

                if (b == 1 && c == 1) update(a, 1, 1, n - 1, 1);

                if (b == 1 && c == 2) update(a, 2, 1, n - 1, 1);

                if (b == 2 && c == 1) update(a, 3, 1, n - 1, 1);

                if (b == 2 && c == 2) update(a, 4, 1, n - 1, 1);

            }

        }

    }

    return 0;

}