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
", 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
"); 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;
}