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  • 2016 计蒜之道 复赛 百度地图的实时路况

    巧妙的解法。

    首先直接限定每一个点不走跑$floyd$是$n^4$的,会超时,我们在$floyd$的时候加入分治思想,用$solve(l, r)$表示不考虑$[l, r]$区间的点所能得到的最短路,然后每一层处理一下就好了,直到最后一层就更新到答案中去。

    时间复杂度$O(n^3logn)$。

    我的代码在$Windows$下一运行就爆栈。

    Code:

    #include <cstdio>
    #include <cstring>
    using namespace std;
    typedef long long ll;
    
    const int N = 305;
    const ll inf = 0x3f3f3f3f3f3f3f3f;
    
    int n;
    ll ans = 0LL;
    
    struct Matrix {
        ll s[N][N];
        
        inline void init() {
            memset(s, 0x3f, sizeof(s));    
        }
        
        inline ll* operator [] (const int now) {
            return s[now];
        }
        
    };
    
    template <typename T>
    inline void read(T &X) {
        X = 0; char ch = 0; T op = 1;
        for(; ch > '9' || ch < '0'; ch = getchar())
            if(ch == '-') op = -1;
        for(; ch >= '0' && ch <= '9'; ch = getchar())
            X = (X << 3) + (X << 1) + ch - 48;
        X *= op;
    }
    
    template <typename T>
    inline void chkMin(T &x, T y) {
        if(y < x) x = y;
    }
    
    void solve(int l, int r, Matrix now) {
        if(l == r) {
            for(int i = 1; i <= n; i++)
                for(int j = 1; j <= n; j++) 
                    if(i != l && j != l) 
                        ans += (now[i][j] == inf) ? -1 : now[i][j]; 
            return;
        }
        
        Matrix tmp = now; 
        int mid = ((l + r) >> 1);
        
        for(int k = mid + 1; k <= r; k++)
            for(int i = 1; i <= n; i++)
                for(int j = 1; j <= n; j++)
                    chkMin(tmp[i][j], tmp[i][k] + tmp[k][j]);
        solve(l, mid, tmp);
        
        tmp = now;
        for(int k = l; k <= mid; k++)
            for(int i = 1; i <= n; i++)
                for(int j = 1; j <= n; j++)
                    chkMin(tmp[i][j], tmp[i][k] + tmp[k][j]);
        solve(mid + 1, r, tmp);
    }
    
    int main() {
        read(n);
        
        Matrix a;
        for(int i = 1; i <= n; i++) 
            for(int j = 1; j <= n; j++) {
                read(a[i][j]);
                if(a[i][j] == -1) a[i][j] = inf;
            }
        
    /*    printf("
    ");
        for(int i = 1; i <= n; i++, printf("
    "))
            for(int j = 1; j <= n; j++)
                printf("%lld ", a[i][j]);   */
        
        solve(1, n, a);
        
        printf("%lld
    ", ans);
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/CzxingcHen/p/9842913.html
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