codeforces round #416 div2

A:暴力模拟

#include<bits/stdc++.h>
using namespace std;
int a, b;
int main()
{
    scanf("%d%d", &a, &b);
    int delta = 1, x = 0;
    while(1)
    {
        if(x == 0)
        {
            if(a < delta)
            {
                puts("Vladik");
                return 0;
            }
            a -= delta; ++delta;
            x ^= 1;
        }
        else
        {
            if(b < delta)
            {
                puts("Valera");
                return 0;
            }
            b -= delta; ++delta;
            x ^= 1;
        }
    }
    return 0;
}

B:先写了个sort竟然pp了，过了十分钟感觉不太好，又写了个基数排序

#include<bits/stdc++.h>
using namespace std;
const int N = 20010;
int n, m;
int p[N], a[N], b[N];
bool cp(int i, int j) { return i < j; }
int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i) scanf("%d", &p[i]);
    while(m--)
    {
        int l, r, x, temp; scanf("%d%d%d", &l, &r, &x);
        for(int i = l; i <= r; ++i) a[i] = p[i], b[p[i]] = 1;
        temp = p[x];
        int pos1 = l, pos2 = 1;
        while(pos1 <= r)
        {
            while(!b[pos2]) ++pos2; b[pos2] = 0;
            a[pos1] = pos2; ++pos1;
        }
        if(a[x] == temp) puts("Yes"); else puts("No");
    }
    return 0;
}

C:dp什么的滚吧。。。dp[i]表示到i的最大值，那么我们枚举i-1每个数选不选，转移一下即可

#include<bits/stdc++.h>
using namespace std;
const int N = 5010;
int n;
int a[N], l[N], r[N], f[N], vis[N];
int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
    {
        scanf("%d", &a[i]);
        r[a[i]] = i;
        if(!l[a[i]]) l[a[i]] = i;
    }
    for(int i = 1; i <= n; ++i)
    {
        f[i] = max(f[i], f[i - 1]); int cur = 0;
        memset(vis, 0, sizeof(vis));
        int last = 1 << 29;
        for(int j = i; j >= 0; --j)
        {
            if(last == j) f[i] = max(f[i], f[j] + cur);
            if(r[a[j]] > i) break;
            if(!vis[a[j]])
            {
                cur ^= a[j];
                vis[a[j]] = 1;
                last = min(last, l[a[j]] - 1);
            }
        }
    }
    printf("%d\n", f[n]);
    return 0;
}

D:没看懂题目

E：很不错的一道题 我们用线段树维护每一层的值和并查集 然后查询的时候把并查集复原合并 太巧妙了 这个并查集复原想了很长时间 思路是这个样子的：因为对于两块东西 原来是不相连的

但是build完和query完并查集都会变 那么我们在build的时候用并查集维护每个节点的连通性 保存两侧的当前并查集中的根是谁。 查询的时候把根的fa赋成自己 因为这个fa代表了原来的连通块

然后之后的节点合并改变了根 那么我们想要回到原来的状态把根赋成单独的一块就行了 然后要注意更新ret要在外面更新 因为可能下面和上面会连接 否则并查集不能及时地更新

#include<bits/stdc++.h>
using namespace std;
#define id(i, j) (i - 1) * m + j
const int N = 1000010;
struct data {
    int sum, l[12], r[12], L, R;
} tree[N];
int n, m, q;
int fa[N], a[12][N];
int find(int x)
{
    return fa[x] == x ? x : fa[x] = find(fa[x]);//fa[x] = find(fa[x]);
}
void connect(int x, int y)
{
    int a = find(x), b = find(y);
    if(a == b) return;
    fa[a] = b;
}
data merge(data x, data y, int mid)
{
    if(!x.sum) return y;
    if(!y.sum) return x;
    data ret;
    ret.sum = x.sum + y.sum;
    for(int i = 1; i <= n; ++i)
    {
        fa[x.l[i]] = x.l[i];
        fa[x.r[i]] = x.r[i];
        fa[y.l[i]] = y.l[i];
        fa[y.r[i]] = y.r[i];
     }
    for(int i = 1; i <= n; ++i)
        if(a[i][mid] == a[i][mid + 1])
        {
            if(find(x.r[i]) != find(y.l[i])) connect(x.r[i], y.l[i]), --ret.sum;
        }
    for(int i = 1; i <= n; ++i)
    {
        ret.l[i] = find(x.l[i]);
        ret.r[i] = find(y.r[i]);
    }
    ret.L = x.L; ret.R = y.R;
    return ret;
}
void build(int l, int r, int x)
{
    if(l == r)
    {
        for(int i = 1; i <= n; ++i)
            if(a[i][l] == a[i - 1][l])
            {
                fa[id(i, l)] = tree[x].l[i] = tree[x].r[i] = tree[x].l[i - 1];
            }
            else fa[id(i, l)] = tree[x].l[i] = tree[x].r[i] = id(i, l), ++tree[x].sum;
        tree[x].L = l; tree[x].R = r;
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, x << 1); build(mid + 1, r, x << 1 | 1);
    tree[x] = merge(tree[x << 1], tree[x << 1 | 1], mid);
}
data query(int l, int r, int x, int a, int b)
{
    if(l > b || r < a) return tree[0];
    if(l >= a && r <= b)
    {
        for(int i = 1; i <= n; ++i)
            fa[id(i, l)] = tree[x].l[i], fa[id(i, r)] = tree[x].r[i];
        return tree[x];
    }
    int mid = (l + r) >> 1;
    data L = query(l, mid, x << 1, a, b);
    data R = query(mid + 1, r, x << 1 | 1, a, b);
    return merge(L, R, mid);
}
int main()
{
    scanf("%d%d%d", &n, &m, &q);
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) scanf("%d", &a[i][j]);
    build(1, m, 1);
    while(q--)
    {
        int l, r; scanf("%d%d", &l, &r);
        printf("%d\n", query(1, m, 1, l, r).sum);
    }
    return 0;
}