湖南集训DAY3

思路:

 1 #include <cstdio>
 2 #include <cctype>
 3 #include <cstring>
 4 
 5 const int MAXN=1000010;
 6 
 7 int n,k;
 8 
 9 char s[MAXN];
10 
11 int hh() {
12     freopen("trans.in","r",stdin);
13     freopen("trans.out","w",stdout);
14     while(scanf("%d%d",&n,&k)!=EOF) {
15         scanf("%s",s+1);
16         for(int i=1; i<=n; ++i) {
17             if(k && s[i]=='2' && s[i+1]=='3') {
18                 if(i&1) s[i+1]='2';
19                 else s[i]='3';
20                 --k;--i;
21             }
22             else if(s[i]=='2' && s[i+1]=='2' && s[i+2]=='3' && (i&1)) k%=2; 
23         }
24         for(int i=1;i<=n;++i) printf("%c",s[i]);
25         printf("\n");
26     }
27     return 0;
28 }
29 
30 int sb=hh();
31 int main(int argc,char**argv) {;}
代码

思路:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int n, a[205][5], m, x, y, f[205][10005][5] = {{{0}}};
 4 int g[10005][5][5] = {{{0}}}, maxi, ans = 0;
 5 bool flag[205][4] = {{false}};
 6 int Main()
 7 {
 8     freopen("snakevsblock.in","r",stdin);
 9     freopen("snakevsblock.out","w",stdout);
10     scanf("%d", &n);
11     for (int i = 1; i <= n; i++)
12         for (int j = 0; j < 5; j++)
13             scanf("%d", &a[i][j]);
14     scanf("%d", &m);
15     for (int i = 1; i <= m; i++)
16     {
17         scanf("%d%d", &x, &y);
18         flag[x][y - 1] = true;
19     }
20     memset(f, -0x7f7f7f, sizeof f);
21     f[0][4][2] = 0;
22     maxi = n * 50;
23     for (int i = 1; i <= n; i++)
24     {
25         memset(g, -0x7f7f7f, sizeof g);
26         for (int j = 0; j <= maxi; j++)
27             for (int k = 0; k < 5; k++)
28                 if (j - a[i][k] >= 0 && j - a[i][k] <= maxi)
29                     f[i][j][k] = g[j][k][k] = f[i - 1][j - a[i][k]][k] + max(-a[i][k], 0);
30         for (int l = 1; l <= 4; l++)
31             for (int j = 0, k = j + l; k < 5; j++, k++)
32                 for (int v = 0, val; v <= maxi; v++)
33                 {
34                     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);
35                     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));
36                     for (int to = j; to <= k; to++) f[i][v][to] = max(f[i][v][to], g[v][j][k]);
37                 }
38     }
39     for (int l = 0; l <= n; l++)
40         for (int i = 0; i <= maxi; i++)
41             for (int j = 0; j < 5; j++)
42                 ans = max(ans, f[l][i][j]);
43     printf("%d\n", ans);
44     return 0;
45 }
46 int dawn = Main ();
47 int main (int argc, char *argv[]) {;}
代码

 

思路:LCA求每一条链的交集 50分

正解

 

 

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<vector>
 6 #define MAXN 100005
 7 #define pb push_back
 8 using namespace std;
 9 
10 struct Edge{ int to,next; }e[MAXN << 1];
11 int head[MAXN],tot,n,m,dep[MAXN],fa[MAXN][20],Ans;
12 vector<int> vec[300000];
13 
14 inline void read(int &x){ 
15     x=0; register char c = getchar();
16     while(c > '9' || c < '0') c = getchar();
17     while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
18 }
19 
20 inline void Add_Edge(int u,int v) {
21     e[++tot].to = v,e[tot].next = head[u],head[u] = tot;
22     e[++tot].to = u,e[tot].next = head[v],head[v] = tot;
23 }
24 
25 void DFS(int u,int fat,int deepth) {
26     fa[u][0] = fat,dep[u] = deepth;
27     for(int i=head[u]; i; i=e[i].next) {
28         int v = e[i].to;
29         if(v == fat) continue;
30         DFS(v,u,deepth + 1);
31     }
32 }
33 
34 void DP() {
35     for(int j=1; j<=18; ++j)
36         for(int i=1; i<=n; ++i)
37             fa[i][j] = fa[fa[i][j-1]][j-1];
38 }
39 
40 void LCA(int u,int v) {
41     if(dep[u] < dep[v]) swap(u,v);
42     while(dep[u] > dep[v]){ u = fa[u][0],vec[Ans].pb(u); }
43     while(u != v){
44         u = fa[u][0], v = fa[v][0];
45         vec[Ans].pb(u);
46         if(u != v) vec[Ans].pb(v);
47     }
48 }
49 
50 void Megre() {
51     vector<int> temp;
52     for(int i=1; i<Ans; ++i) {
53         for(int j=0; j<vec[Ans].size(); ++j)
54             for(int k=0; k<vec[i].size(); ++k)
55                 if(vec[i][k] == vec[Ans][j]) temp.pb(vec[Ans][j]);
56         if(!temp.empty()){
57             vec[i].clear(),vec[Ans].clear(),Ans--;
58             for(int k=0; k<temp.size(); ++k)
59                 vec[i].pb(temp[k]);
60             break;
61         }
62     }
63 }
64 
65 int main(int agrc,char *argv[]) {
66     freopen("ping.in","r",stdin);
67     freopen("ping.out","w",stdout);
68     read(n),read(m); register int i; int u,v,k;
69     for(i=1; i<=m; ++i)
70         read(u),read(v),Add_Edge(u,v);
71     DFS(1,0,-1);DP();
72     read(k);
73     for(i=1; i<=k; ++i) {
74         read(u),read(v);
75         ++Ans;
76         vec[Ans].pb(u);
77         if(u != v) vec[Ans].pb(v);
78         LCA(u,v);
79         Megre();
80     }
81     printf("%d\n",Ans);
82     for(int i=1; i<=Ans; ++i)
83         printf("%d ",vec[i][0]);
84     fclose(stdin); fclose(stdout);
85     return 0;
86 }
50分代码
  1 #include <cstdio>
  2 #include<algorithm>
  3 using namespace std;
  4 int n, m, P, u, v, to[200005], nxt[200005], p[100005], deep[100005], q[500005][2];
  5 int son[100005], fa[100005], size[100005], top[100005], dfsx[100005], cnt, ans;
  6 int lca[500005], Q[500005], lr[100005][2];
  7 int sta[100005];
  8 bool flag[400005];
  9 void dfs(int x) {
 10     size[x] = 1;
 11     for (int i = p[x]; i != -1; i = nxt[i])
 12         if (to[i] != fa[x]) {
 13             fa[to[i]] = x;
 14             deep[to[i]] = deep[x] + 1;
 15             dfs(to[i]);
 16             if (son[x] == -1 || size[to[i]] > size[son[x]]) son[x] = to[i];
 17             size[x] += size[to[i]];
 18         }
 19 }
 20 void dfs1(int x) {
 21     dfsx[x] = ++cnt;
 22     if (son[x] != -1) top[son[x]] = top[x], dfs1(son[x]);
 23     for (int i = p[x]; i != -1; i = nxt[i])
 24         if (to[i] != fa[x] && to[i] != son[x])
 25             top[to[i]] = to[i], dfs1(to[i]);
 26 }
 27 int findlca(int x, int y) {
 28     while (1) {
 29         if (top[x] == top[y]) return deep[x] > deep[y]? y : x;
 30         if (deep[top[x]] > deep[top[y]]) x = fa[top[x]];
 31         else y = fa[top[y]];
 32     }
 33 }
 34 bool query(int x, int l, int r, int ll, int rr) {
 35     if (l == ll && r == rr) return flag[x];
 36     int mid = (l + r) >> 1, L = x << 1, R = L | 1;
 37     if (rr <= mid) return query(L, l, mid, ll, rr);
 38     else if (ll > mid) return query(R, mid + 1, r, ll, rr);
 39     else return query(L, l, mid, ll, mid) | query(R, mid + 1, r, mid + 1, rr);
 40 }
 41 void modify(int x, int l, int r, int to) {
 42     flag[x] = true;
 43     if (l == r) return;
 44     int mid = (l + r) >> 1, L = x << 1, R = L | 1;
 45     if (to <= mid) modify(L, l, mid, to);
 46     else modify(R, mid + 1, r, to);
 47 }
 48 bool Query(int x, int y) {
 49     while (1) {
 50         if (top[x] == top[y]) {
 51             if (deep[x] < deep[y]) return query(1, 1, cnt, dfsx[x], dfsx[y]);
 52             else return query(1, 1, cnt, dfsx[y], dfsx[x]);
 53         }
 54         if (deep[top[x]] > deep[top[y]])
 55             if (query(1, 1, cnt, dfsx[top[x]], dfsx[x])) return true;
 56             else x = fa[top[x]];
 57         else {
 58             if (query(1, 1, cnt, dfsx[top[y]], dfsx[y])) return true;
 59             else y = fa[top[y]];
 60         }
 61     }
 62 }
 63 void work(int x) {
 64     for (int i = p[x]; i != -1; i = nxt[i])
 65         if (to[i] != fa[x])
 66             work(to[i]);
 67     for (int i = lr[x][0]; i <= lr[x][1]; i++)
 68         if (!Query(q[Q[i]][0], q[Q[i]][1])) {
 69             modify(1, 1, cnt, dfsx[x]);
 70             sta[++ans] = x;
 71             return;
 72         }
 73 }
 74 bool cmp(int x, int y) {
 75     return lca[x] < lca[y];
 76 }
 77 int main() {
 78     freopen("ping.in","r",stdin);
 79     freopen("ping.out","w",stdout);
 80     scanf("%d%d", &n, &m);
 81     for (int i = 0; i <= n; i++) p[i] = son[i] = -1, top[i] = size[i] = fa[i] = deep[i] = 0;
 82     for (int i = 1; i <= n * 4; i++) flag[i] = false;
 83     for (int i = 1; i <= m; i++) {
 84         scanf("%d%d", &u, &v);
 85         u--, v--;
 86         to[i * 2 - 1] = v;
 87         nxt[i * 2 - 1] = p[u];
 88         p[u] = i * 2 - 1;
 89         to[i * 2] = u;
 90         nxt[i * 2] = p[v];
 91         p[v] = i * 2;
 92     }
 93     deep[0] = 1;
 94     dfs(0);
 95     cnt = 0;
 96     dfs1(0);
 97     scanf("%d", &P);
 98     for (int i = 1; i <= P; i++) {
 99         scanf("%d%d", &u, &v);
100         u--, v--;
101         q[i][0] = u, q[i][1] = v;
102         lca[i] = findlca(u, v);
103         Q[i] = i;
104     }
105     sort(Q + 1, Q + P + 1, cmp);
106     for (int i = 0; i <= n; i++) lr[i][0] = P + 1, lr[i][1] = 0;
107     for (int i = 1; i <= P; i++) {
108         lr[lca[Q[i]]][0] = min(lr[lca[Q[i]]][0], i);
109         lr[lca[Q[i]]][1] = max(lr[lca[Q[i]]][1], i);
110     }
111     ans = 0;
112     work(0);
113     printf("%d\n", ans);
114     for (int i = 1; i <= ans; i++) printf("%d ", sta[i] + 1);
115     return 0;
116 }
std

 

posted @ 2017-10-20 15:26  拿叉插猹哈  阅读(114)  评论(0编辑  收藏  举报