• 无题4


    第一题:O (n*m) 对于查询矩阵xl, yl, xr, yr;

    可以变为 sum(x, i) * (i, y)   xl <= x <= xr, yl <= y <= yr, 1 <= i <= n;

    用一个乘法分配率+前缀和 变为sum ( a[xl - xr][i] * b[i][yl - yr]);

    #include<bits/stdc++.h>
    using namespace std;
    #define ll long long
    const int M = 2005;
    ll sum1[M][M], sum2[M][M], b[M][M], a[M][M], tot;
    
    int main(){
        freopen("matrix.in","r",stdin);
        freopen("matrix.out","w",stdout);
        int n, m, x;
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++){
                scanf("%lld", &a[i][j]);
                sum1[i][j] = sum1[i - 1][j] + a[i][j];
            }
                
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++){
                scanf("%lld", &b[i][j]);
                sum2[i][j] = sum2[i][j - 1] + b[i][j];
            }
            
        int xl, yl, xr, yr;
        while(m--){
            ll ans = 0;
            scanf("%d%d%d%d", &xl, &yl, &xr, &yr);
            if(xl > xr) swap(xl, xr);
            if(yl > yr) swap(yl, yr);
            for(int j = 1; j <= n; j++)
                ans += (sum1[xr][j] - sum1[xl - 1][j]) * (sum2[j][yr] - sum2[j][yl - 1]);
            printf("%lld
    ",ans);
        }
    }
    View Code

    第二题:X轴,Y轴可以分开处理,就变成了一个中位数问题(证明蓝书上有);因为肯定有一个中位数即一个点存在,我们就可以枚举这个中位数,X,Y各枚举一个,取距离的前i小;

    O(n^4 * logn)

    #include<bits/stdc++.h>
    using namespace std;
    
    const int M = 55, inf = 2e9;
    int dis[M];
    struct Point{int x, y;}a[M];
    inline int ab(int a){return a >= 0 ? a : -a;}
    inline int get(int x, int y, int t){
        return ab(x - a[t].x) + ab(y - a[t].y);
    }
    
    
    int main(){
        freopen("tower.in","r",stdin);
        freopen("tower.out","w",stdout);
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++)scanf("%d%d", &a[i].x, &a[i].y);
        for(int cnt = 1; cnt <= n; cnt++){
            int ans = inf;
            for(int i = 1; i <= n; i++){
                for(int j = 1; j <= n; j++){
                    int xx = a[i].x, yy = a[j].y;
                    int tmp = 0;
                    for(int p = 1; p <= n; p++)
                        dis[p] = get(xx, yy, p);
                    sort(dis + 1, dis + 1 + n);
                    for(int p = 1; p <= cnt; p++)
                    tmp += dis[p];
                    ans = min(ans, tmp);    
                }
                
            }
            printf("%d
    ", ans);
            
        }
        return 0;
    } 
    View Code

    第三题:树形DP+高精度;

    dp[u][0/1]表示u的子树中的最大匹配,g[u][0/1]表示方案数;

    更新dp, 对于u和下面连的情况就暴力枚举和哪一个连;

    dp是用那种更新的,g就用那种更新,乘法原理, 如果dp值一样,就加法原理

    下面是无高精的code

    #include<bits/stdc++.h>
    using namespace std;
    #define ll long long
    const int M = 1e3 + 10;
    ll g[M][2];
    int h[M], dp[M][2], tot;
    
    struct edge{int v, nxt;}G[M << 1];
    void add(int u, int v){G[++tot].v = v; G[tot].nxt = h[u]; h[u] = tot;}
    void dfs(int u, int f){
        int child = 0, ans = 0;
        ll tmp1 = 1, opt = 0;
        for(int i = h[u]; i; i = G[i].nxt){
            int v = G[i].v;
            if(v == f)continue;
            dfs(v, u);
            child++;
            if(dp[v][0] > dp[v][1]) tmp1 *= g[v][0];
            else if(dp[v][0] == dp[v][1]) tmp1 *= (g[v][0] + g[v][1]);
            else tmp1 *= g[v][1];
            dp[u][0] += max(dp[v][0], dp[v][1]);        
        }
        g[u][0] = tmp1;
        
        for(int i = h[u]; i; i = G[i].nxt){
            int v = G[i].v;
            if(v == f)continue;
            int d = dp[v][0]; 
            ll tmp2 = g[v][0];
            for(int j = h[u]; j; j = G[j].nxt){
                int vv = G[j].v;
                if(vv != f && vv != v){
                    if(dp[vv][0] > dp[vv][1]) tmp2 *= g[vv][0];
                    else if(dp[vv][0] == dp[vv][1]) tmp2 *= (g[vv][0] + g[vv][1]);
                    else tmp2 *= g[vv][1];    
                    d += max(dp[vv][0], dp[vv][1]);
                }
                
            }
            if(d > ans)ans = d, opt = tmp2;
            else if(d == ans) opt += tmp2;
        }
        if(child) dp[u][1] = ans + 1, g[u][1] = opt;
        //printf("%d %d %d %lld %lld
    ", u, dp[u][0], dp[u][1], g[u][0], g[u][1]);
        
    }
    
    int main(){
        freopen("tree.in","r", stdin);
        freopen("tree.out","w",stdout);
        int n, u, x, v;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++){
            scanf("%d%d",&u, &x);
            while(x--) {
                scanf("%d", &v);
                add(u, v);
            }
        }
        dfs(1, 0);
        ll ret = 0;
        if(dp[1][0] >= dp[1][1]) ret += g[1][0];
        if(dp[1][1] >= dp[1][0]) ret += g[1][1];
        printf("%d
    %lld
    ", max(dp[1][0], dp[1][1]), ret);
    } 
    View Code
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  • 原文地址:https://www.cnblogs.com/EdSheeran/p/9675319.html
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