湖南集训day3
Posted 安
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了湖南集训day3相关的知识，希望对你有一定的参考价值。
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=0; i<n; i++) { if(!m) break; if(a[i]==\'2\' && a[i+1]==\'2\' && a[i+2]==\'3\') { if(i%2==0) { if(m%2==0) break; else { a[i+1]=\'3\'; break; } } else { m--; a[i+2]=\'2\'; continue; } } else if(a[i]==\'2\' && a[i+1]==\'3\' && a[i+2] ==\'3\') { if(i%2==1) { a[i]=\'3\'; m--; continue; } else { if(m%2==0) break; else { a[i+1]=\'2\'; break; } } } else if(a[i]==\'2\' && a[i+1]==\'3\') { m--; if(i%2) a[i]=\'3\'; else a[i+1]=\'2\'; } } printf("%s",a);printf("\\n"); } fclose(stdin);fclose(stdout); return 0; }
/* 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][5],f[N][M][5],g[M][5][5]; bool flag[N][4]; int n,m,maxi,ans,val; inline int read() { int x=0,f=1;char c=getchar(); while(c>\'9\'||c<\'0\'){if(c==\'-\')f=-1;c=getchar();} while(c>=\'0\'&&c<=\'9\'){x=x*10+c-\'0\';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=1;i<=n;i++) for(int j=0;j<5;j++) a[i][j]=read(); m=read(); for(int i=1;i<=m;i++) { x=read();y=read(); flag[x][y-1]=1; } memset(f,-0x7f7f7f,sizeof f); f[0][4][2]=0;maxi=n*50; for(int i=1;i<=n;i++) { memset(g,-0x7f7f7f,sizeof g); for(int j=0;j<=maxi;j++) for(int k=0;k<5;k++) if(j-a[i][k]>=0 && j-a[i][k]<=maxi) f[i][j][k]=g[j][k][k]=f[i-1][j-a[i][k]][k]+max(-a[i][k],0); for(int l=1;l<=4;l++) for(int j=0,k=j+l;k<5;j++,k++) for(int v=0;v<=maxi;v++) { if(!flag[i][j] && (val=v-a[i][j])>=0 && val<=maxi) g[v][j][k] = g[val][j + 1][k] + max(-a[i][j], 0);//g[v][j][k]=max(g[v][j][k],g[val][j+1][k]+max(-a[i][j],0)); if(!flag[i][k-1]&& (val=v-a[i][k])>=0 && val<=maxi) g[v][j][k]=max(g[v][j][k],g[val][j][k-1]+max(-a[i][k],0)); for(int to=j;to<=k;to++) f[i][v][to]=max(f[i][v][to],g[v][j][k]); } } for(int i=0;i<=n;i++) for(int j=0;j<=maxi;j++) for(int k=0;k<5;k++) ans=max(f[i][j][k],ans); printf("%d\\n",ans); return 0; }
、
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;
}
以上是关于湖南集训day3的主要内容，如果未能解决你的问题，请参考以下文章