• BZOJ1898: [Zjoi2005]Swamp 沼泽鳄鱼(矩阵快速幂)


    题意

    题目链接

    Sol

    不难发现吃人鱼的运动每(12s)一个周期

    所以暴力建12个矩阵,放在一起快速幂即可

    最后余下的部分暴力乘

    #include<bits/stdc++.h>
    using namespace std;
    const int MAXN = 52, mod = 10000;
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N, M, S, T, K, mp[MAXN][MAXN], pos[MAXN];
    int add(int x, int y) {
        if(x + y < 0) return x + y + mod;
        return x + y >= mod ? x + y - mod : x + y;
    }
    int add2(int &x, int y) {
        if(x + y < 0) x = x + y + mod;
        else x = (x + y >= mod ? x + y - mod : x + y);
    }
    int mul(int x, int y) {
        return 1ll * x * y % mod;
    }
    struct Ma {
        int m[MAXN][MAXN];
        Ma() {
            memset(m, 0, sizeof(m));
        }
        Ma operator * (const Ma &rhs) const {
            Ma ans;
            for(int k = 1; k <= N; k++)
                for(int i = 1; i <= N; i++)
                    for(int j = 1; j <= N; j++) 
                        add2(ans.m[i][j], mul(m[i][k], rhs.m[k][j]));
            return ans;
        }
    }a[15], bg;
    Ma MatrixFp(Ma a, int p) {
        Ma base;
        for(int i = 1; i <= N; i++) base.m[i][i] = 1;
        while(p) {
            if(p & 1) base = base * a;
            a = a * a; p >>= 1;
        }
        return base;
    }
    int main() {
        N = read(); M = read(); S = read() + 1; T = read() + 1; K = read();
        for(int i = 1; i <= M; i++) {int x = read() + 1, y = read() + 1; bg.m[x][y]++; bg.m[y][x]++;}
        int fn = read();
        for(int i = 1; i <= 12; i++) a[i] = bg;
        for(int i = 1; i <= fn; i++) {
            int num = read();
            for(int j = 1; j <= num; j++) pos[j] = read() + 1;
            for(int j = 1; j <= 12; j++)
                for(int k = 1; k <= N; k++)
                    a[j].m[k][pos[j % num + 1]] = 0;
        }
        Ma res = a[1];
        //for(int i = 1; i <= N; i++) res.m[i][i] = 1;
        for(int i = 2; i <= 12; i++) res = res * a[i];
        res = MatrixFp(res, K / 12);
        for(int i = 1; i <= K % 12; i++) res = res * a[i];
        printf("%d
    ", res.m[S][T]);
        return 0;
    }
    /*
    6 8 1 5 333
    0 2
    2 1
    1 0
    0 5
    5 1
    1 4
    4 3
    3 5
    3
    3 0 5 1
    2 1 2
    4 1 2 3 4 
    */
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10028973.html
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