Codeforces Round #684 (Div. 2)

A

讨论三种情况，不换/全换成0/全换成1 ,取一个花费最小值

#include <bits/stdc++.h>
using namespace std;
const int N = 1000 + 20; 

int n, c0, c1, h;
char str[N];

int main()
{
	int T;
	scanf("%d", &T);
	while(T -- )
	{
		scanf("%d%d%d%d%s", &n, &c0, &c1, &h, str);
		int a = 0, b = 0;
		for(int i = 0; i < n; ++ i)
			if(str[i] == '0') a ++;
			else b ++;
		int res = 1e9;
		res = min(res, n * c0 + h * b);
		res = min(res, n * c1 + h * a);
		res = min(res, c0 * a + c1 * b);
		printf("%d\n", res);
	}
	return 0;
}

B

排序之后，从\(n \times k - \dfrac{n}{2}\)开始取，每隔\(\dfrac{n}{2}\)取一个，取\(k\)个即可

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1000 + 20;

int n, k, a[N * N];

int main()
{
	int T;
	scanf("%d", &T);
	while(T -- )
	{
		scanf("%d%d", &n, &k);
		for(int i = 1; i <= n * k; ++ i) scanf("%d", &a[i]);
		sort(a + 1, a + n * k + 1);
		LL res = 0;
		for(int i = n * k - n / 2, j = 1; j <= k; i -= n / 2 + 1, ++ j)
			res += a[i];
		printf("%lld\n", res);
	}
	return 0;
}

C1/C2

按行处理，处理前\(n - 2\)行，最后\(2\)行按列处理，处理到\(m - 2\)列，最后\(4\)个格子讨论即可.

#include <bits/stdc++.h>
using namespace std;
const int N = 100 + 5;

int n, m;
int a[N][N];

struct zt
{
    int a, b, c, d, e, f;
};
vector<zt> vec;

int main()
{
	int T;
	scanf("%d",&T);
	while(T --)
	{
        vec.clear();
		scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; ++ i)
            for(int j = 1; j <= m; ++ j)
                    scanf("%1d", &a[i][j]);
        for(int i = 1; i <= n - 2; ++ i)
            for(int j = 1; j <= m; ++ j)
                if(a[i][j])
                {
                    vec.push_back((zt){i, j, i + 1, j, i + 1, j < m ? j + 1 : j - 1});
                    a[i][j] ^= 1; a[i + 1][j] ^= 1; a[i + 1][j < m ? j + 1 : j - 1] ^= 1;
                }
        for(int j = 1; j <= m - 2; ++ j)
            for(int i = n - 1; i <= n; ++ i)
                if(a[i][j])
                {
                    vec.push_back((zt){i, j, i, j + 1, i < n ? i + 1 : i - 1, j + 1});
                    a[i][j] ^= 1; a[i][j + 1] ^= 1; a[i < n ? i + 1 : i - 1][j + 1] ^= 1;
                }
        int num = a[n - 1][m - 1] + a[n - 1][m] + a[n][m - 1] + a[n][m];
        if(num == 1)
        {
            if(a[n][m])
            {
                vec.push_back((zt){n, m, n, m - 1, n - 1, m - 1});
                vec.push_back((zt){n, m, n, m - 1, n - 1, m});
                vec.push_back((zt){n, m, n - 1, m, n - 1, m - 1});
            }
            if(a[n - 1][m])
            {
                vec.push_back((zt){n - 1, m, n - 1, m - 1, n, m - 1});
                vec.push_back((zt){n - 1, m, n - 1, m - 1, n, m});
                vec.push_back((zt){n - 1, m, n, m, n, m - 1});
            }
            if(a[n][m - 1])
            {
                vec.push_back((zt){n, m - 1, n - 1, m - 1, n - 1, m});
                vec.push_back((zt){n, m - 1, n - 1, m - 1, n, m});
                vec.push_back((zt){n, m - 1, n, m, n - 1, m});
            }
            if(a[n - 1][m - 1])
            {
                vec.push_back((zt){n - 1, m - 1, n, m - 1, n, m});
                vec.push_back((zt){n - 1, m - 1, n, m - 1, n - 1, m});
                vec.push_back((zt){n - 1, m - 1, n - 1, m, n, m});
            }
        }
        if(num == 2)
        {
            if(a[n - 1][m - 1] && a[n - 1][m])
            {
                vec.push_back((zt){n - 1, m - 1, n, m - 1, n, m});
                vec.push_back((zt){n, m - 1, n, m, n - 1, m});
            }
            if(a[n][m - 1] && a[n][m])
            {
                vec.push_back((zt){n, m - 1, n - 1, m - 1, n - 1, m});
                vec.push_back((zt){n, m, n - 1, m, n - 1, m - 1});
            }
            if(a[n - 1][m - 1] && a[n][m - 1])
            {
                vec.push_back((zt){n - 1, m - 1, n - 1, m, n, m});
                vec.push_back((zt){n, m - 1, n - 1, m, n, m});
            }
            if(a[n - 1][m] && a[n][m])
            {
                vec.push_back((zt){n - 1, m, n - 1, m - 1, n, m - 1});
                vec.push_back((zt){n, m, n, m - 1, n - 1, m - 1});
            }
            if(a[n - 1][m - 1] && a[n][m])
            {
                vec.push_back((zt){n - 1, m - 1, n - 1, m, n, m - 1});
                vec.push_back((zt){n, m, n - 1, m, n, m - 1});
            }
            if(a[n][m - 1] && a[n - 1][m])
            {
                vec.push_back((zt){n - 1, m, n - 1, m - 1, n, m});
                vec.push_back((zt){n, m - 1, n - 1, m - 1, n, m});
            }
        }
        if(num == 3)
        {
            vector<int> tmp;
            if(a[n][m]) tmp.push_back(n), tmp.push_back(m);
            if(a[n - 1][m - 1]) tmp.push_back(n - 1), tmp.push_back(m - 1);
            if(a[n - 1][m]) tmp.push_back(n - 1), tmp.push_back(m);
            if(a[n][m - 1]) tmp.push_back(n), tmp.push_back(m - 1);
            vec.push_back((zt){tmp[0], tmp[1], tmp[2], tmp[3], tmp[4], tmp[5]});
        }
        if(num == 4)
        {
            vec.push_back((zt){n - 1, m - 1, n - 1, m, n, m - 1});
            vec.push_back((zt){n, m, n, m - 1, n - 1, m - 1});
            vec.push_back((zt){n, m, n, m - 1, n - 1, m});
            vec.push_back((zt){n, m, n - 1, m, n - 1, m - 1});
        }
        printf("%d\n", vec.size());
        for(int i = 0; i < vec.size(); ++ i)
            printf("%d %d %d %d %d %d\n", vec[i].a, vec[i].b, vec[i].c, vec[i].d, vec[i].e, vec[i].f);
	}
	return 0;
}

E

线段树，维护一个min和max和区间和，利用单调不增的性质进行修改和查询

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;

int n, m, w[N];
struct Node
{
	int l, r;
	LL maxv, minv, sum, lazy;
}tr[N * 4];

void pushup(int u)
{
	tr[u].maxv = max(tr[u << 1].maxv, tr[u << 1 | 1].maxv);
	tr[u].minv = min(tr[u << 1].minv, tr[u << 1 | 1].minv);
	tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}

void pushdown(int u)
{
	Node &root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
	if(root.lazy)
	{
		left.maxv = right.maxv = root.lazy;
		left.minv = right.minv = root.lazy;
		left.sum = (left.r - left.l + 1) * root.lazy;
		right.sum = (right.r - right.l + 1) * root.lazy;
		left.lazy = right.lazy = root.lazy;
		root.lazy = 0;
	}
}

void build(int u, int l, int r)
{
	if(l == r) tr[u] = {l, r, w[r], w[r], w[r], 0};
	else
	{
		tr[u] = {l, r, 0, 0, 0, 0};
		int mid = l + r >> 1;
		build(u << 1, l, mid);
		build(u << 1 | 1, mid + 1, r);
		pushup(u);
	}
}

void modify(int u, int l, int r, int v)
{
	if(tr[u].minv >= v) return;
	if(tr[u].l >= l && tr[u].r <= r && tr[u].maxv < v)
	{
		tr[u].lazy = v;
		tr[u].minv = tr[u].maxv = v;
		tr[u].sum = (LL)(tr[u].r - tr[u].l + 1) * v;
	}
	else
	{
		pushdown(u);
		int mid = tr[u].l + tr[u].r >> 1;
		if(l <= mid) modify(u << 1, l, r, v);
		if(r > mid) modify(u << 1 | 1, l, r, v);
		pushup(u);
	}
}

int query(int u, int l, int r, int &v)
{
	if(tr[u].minv > v) return 0;
	if(tr[u].l >= l && tr[u].r <= r && tr[u].sum <= v)
	{
		v -= tr[u].sum;
		return tr[u].r - tr[u].l + 1;
	}
	else
	{
		pushdown(u);
		int mid = tr[u].l + tr[u].r >> 1;
		int res = 0;
		if(l <= mid) res = query(u << 1, l, r, v);
		if(r > mid) res += query(u << 1 | 1, l, r, v);
		return res;
	}
}

int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; ++ i) scanf("%d", &w[i]);
	build(1, 1, n);
	while(m -- )
	{
		int op, x, y;
		scanf("%d%d%d", &op, &x, &y);
		if(op == 1) modify(1, 1, x, y);
		else printf("%d\n", query(1, x, n, y));
	}
	return 0;
}

2020.11.19