湖南集训day3

难度：☆☆☆☆☆☆☆

此时相望不相闻，愿逐月华流照君

/*
23
233
223
啦啦啦德玛西亚
*/
#include<iostream>
#include<cstdio>
#include<cstring>

#define N 1000007

using namespace std;
int n,m,ans;
char a[N];

int main()
{
    freopen("trans.in","r",stdin);
    freopen("trans.out","w",stdout);
    while(~scanf("%d%d%s",&n,&m,a))
    {
        for(int i=; i<n; i++)
        {
            if(!m) break;
            if(a[i]=='' && a[i+]=='' && a[i+]=='')
            {
                if(i%==)
                {
                    if(m%==) break;
                    else
                    {
                        a[i+]='';
                        break;
                    }
                }
                else
                {
                    m--;
                    a[i+]='';
                    continue;
                }
            }
            else if(a[i]=='' && a[i+]=='' && a[i+]

=='')
            {
                if(i%==)
                {
                    a[i]='';
                    m--;
                    continue;
                }
                else
                {
                    if(m%==) break;
                    else
                    {
                        a[i+]='';
                        break;
                    }
                }
            }
            else if(a[i]=='' && a[i+]=='')
            {
                m--;
                if(i%) a[i]='';
                else a[i+]='';
            }
        }
        printf("%s",a);printf("\n");
    }

    fclose(stdin);fclose(stdout);
    return ;
}

/*
把豆豆和砖块丢在一起dp
f[i][j][k]:到第i行,蛇的长度还剩下j,从k位置出第i行的最大得分
g[j][l][r]:蛇的长度剩下j,从当前行l到r区间内仍为死亡的最大得分
初始化:对于每个i，g[j][k][k]=f[i-1][j-a[i][k]][k]+max(-a[i][k],0) 为初始状态
方程g[j][l][r] = max(g[j-a[i][l]][l+1][r] + max(-a[i][j],0),g[j-a[i][r]][l][r-1]+max(-a[i][k],0));
最后f[i][j][k] = max{g[j][l][r](1≤l≤k ≤r ≤5)}
*/
#include<iostream>
#include<cstdio>
#include<cstring>

#define N 207
#define M 10005

using namespace std;
int a[N][],f[N][M][],g[M][][];
bool flag[N][];
int n,m,maxi,ans,val;

inline int read()
{
    int x=,f=;char c=getchar();
    while(c>''||c<''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x*f;
}

int main()
{
    freopen("snakevsblock.in","r",stdin);
    freopen("snakevsblock.out","w",stdout);
    n=read();int x,y;
    for(int i=;i<=n;i++) for(int j=;j<;j++)
      a[i][j]=read();
    m=read();
    for(int i=;i<=m;i++)
    {
        x=read();y=read();
        flag[x][y-]=;
    }
    memset(f,-0x7f7f7f,sizeof f);
    f[][][]=;maxi=n*;

    for(int i=;i<=n;i++)
    {
        memset(g,-0x7f7f7f,sizeof g);
        for(int j=;j<=maxi;j++)
          for(int k=;k<;k++)
            if(j-a[i][k]>= && j-a[i][k]<=maxi)
              f[i][j][k]=g[j][k][k]=f[i-][j-a[i][k]][k]+max(-a[i][k],);

        for(int l=;l<=;l++)
          for(int j=,k=j+l;k<;j++,k++)
            for(int v=;v<=maxi;v++)
            {
                if(!flag[i][j]  && (val=v-a[i][j])>= && val<=maxi) g[v][j][k] = g[val][j + ][k] + max(-a[i][j], );//g[v][j][k]=max(g[v][j][k],g[val][j+1][k]+max(-a[i][j],0));
                if(!flag[i][k-]&& (val=v-a[i][k])>= && val<=maxi) g[v][j][k]=max(g[v][j][k],g[val][j][k-]+max(-a[i][k],));
                for(int to=j;to<=k;to++) f[i][v][to]=max(f[i][v][to],g[v][j][k]);
            }
    }

    for(int i=;i<=n;i++)
      for(int j=;j<=maxi;j++)
        for(int k=;k<;k++)
          ans=max(f[i][j][k],ans);

    printf("%d\n",ans);
    return ;
}

、

不会不会不会不会不会不会就不会................................................................................................

暂时弃疗

题解：

std：

#include <bits/stdc++.h>
using namespace std;
int n, m, P, u, v, to[200005], nxt[200005], p[100005], deep[100005], q[500005][2];
int son[100005], fa[100005], size[100005], top[100005], dfsx[100005], cnt, ans;
int lca[500005], Q[500005], lr[100005][2];
int sta[100005];
bool flag[400005];
void dfs(int x)
{
	size[x] = 1;
	for (int i = p[x]; i != -1; i = nxt[i])
		if (to[i] != fa[x])
		{
			fa[to[i]] = x;
			deep[to[i]] = deep[x] + 1;
			dfs(to[i]);
			if (son[x] == -1 || size[to[i]] > size[son[x]]) son[x] = to[i];
			size[x] += size[to[i]];
		}
}
void dfs1(int x)
{
	dfsx[x] = ++cnt;
	if (son[x] != -1) top[son[x]] = top[x], dfs1(son[x]);
	for (int i = p[x]; i != -1; i = nxt[i])
		if (to[i] != fa[x] && to[i] != son[x])
			top[to[i]] = to[i], dfs1(to[i]);
}
int findlca(int x, int y)
{
	while (1)
	{
		if (top[x] == top[y]) return deep[x] > deep[y]? y : x;
		if (deep[top[x]] > deep[top[y]]) x = fa[top[x]];
		else y = fa[top[y]];
	}
}
bool query(int x, int l, int r, int ll, int rr)
{
	if (l == ll && r == rr) return flag[x];
	int mid = (l + r) >> 1, L = x << 1, R = L | 1;
	if (rr <= mid) return query(L, l, mid, ll, rr);
	else if (ll > mid) return query(R, mid + 1, r, ll, rr);
	else return query(L, l, mid, ll, mid) | query(R, mid + 1, r, mid + 1, rr);
}
void modify(int x, int l, int r, int to)
{
	flag[x] = true;
	if (l == r) return;
	int mid = (l + r) >> 1, L = x << 1, R = L | 1;
	if (to <= mid) modify(L, l, mid, to);
	else modify(R, mid + 1, r, to);
}
bool Query(int x, int y)
{
	while (1)
	{
		if (top[x] == top[y])
		{
			if (deep[x] < deep[y]) return query(1, 1, cnt, dfsx[x], dfsx[y]);
			else return query(1, 1, cnt, dfsx[y], dfsx[x]);
		}
		if (deep[top[x]] > deep[top[y]])
			if (query(1, 1, cnt, dfsx[top[x]], dfsx[x])) return true;
			else x = fa[top[x]];
		else
		{
			if (query(1, 1, cnt, dfsx[top[y]], dfsx[y])) return true;
			else y = fa[top[y]];
		}
	}
}
void work(int x)
{
	for (int i = p[x]; i != -1; i = nxt[i])
		if (to[i] != fa[x])
			work(to[i]);
	for (int i = lr[x][0]; i <= lr[x][1]; i++)
		if (!Query(q[Q[i]][0], q[Q[i]][1]))
		{
			modify(1, 1, cnt, dfsx[x]);
			sta[++ans] = x;
			return;
		}
}
bool cmp(int x, int y) {return lca[x] < lca[y];}
int main()
{
	freopen("ping.in","r",stdin);
	freopen("ping.out","w",stdout);
	scanf("%d%d", &n, &m);
	for (int i = 0; i <= n; i++) p[i] = son[i] = -1, top[i] = size[i] = fa[i] = deep[i] = 0;
	for (int i = 1; i <= n * 4; i++) flag[i] = false;
	for (int i = 1; i <= m; i++)
	{
		scanf("%d%d", &u, &v);
		u--, v--;
		to[i * 2 - 1] = v;
		nxt[i * 2 - 1] = p[u];
		p[u] = i * 2 - 1;
		to[i * 2] = u;
		nxt[i * 2] = p[v];
		p[v] = i * 2;
	}
	deep[0] = 1;
	dfs(0);
	cnt = 0;
	dfs1(0);
	scanf("%d", &P);
	for (int i = 1; i <= P; i++)
	{
		scanf("%d%d", &u, &v);
		u--, v--;
		q[i][0] = u, q[i][1] = v;
		lca[i] = findlca(u, v);
		Q[i] = i;
	}
	sort(Q + 1, Q + P + 1, cmp);
	for (int i = 0; i <= n; i++) lr[i][0] = P + 1, lr[i][1] = 0;
	for (int i = 1; i <= P; i++)
	{
		lr[lca[Q[i]]][0] = min(lr[lca[Q[i]]][0], i);
		lr[lca[Q[i]]][1] = max(lr[lca[Q[i]]][1], i);
	}
	ans = 0;
	work(0);
	printf("%d\n", ans);
	for (int i = 1; i <= ans; i++) printf("%d ", sta[i] + 1);
	return 0;
}