Codeforces Round #310 (Div. 1) 完整题解

Codeforces Round #310 (Div. 1) 完整题解

A题：看懂题意后，只要找到从1开始连续的长度就行了，其他的套娃都要全拆开。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 2000005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// headint a[maxn];
int n, m;void work()
{int k, t2 = 0;int ans = 0, t = 1;for(int i = 1; i <= m; i++) {scanf("%d", &k);for(int j = 1; j <= k; j++) scanf("%d", &a[j]);if(a[1] == 1) {for(int j = 2; j <= k; j++) {if(a[j] == a[j-1] + 1) t++;else {t2 = 1;break;}}}}ans = (n - t) - (m - 1) + (n - t);printf("%d\n", ans);
}int main()
{while(scanf("%d%d", &n, &m)!=EOF) {work();}return 0;
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B题：贪心，先预处理出n-1条桥的长度，设每条桥的区间为[l,r]，那么可以证明找到比l大的且长度最小的桥是最优的。证明如下：1.设第l[i] > l[i+1] r[i] < r[i+1]，那么先满足第i个桥的条件必然是最优的。2.l[i] < l[i+1]  r[i] < r[i+1]，画个图就知道找比l[i]大的最小值也是最优的。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 200005
#define maxm 2000005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// headstruct node
{LL l, r;int id;
}p[maxn];set<pair<LL, int> > s;
set<pair<LL, int> >::iterator it;
LL l[maxn];
LL r[maxn];
int ans[maxn];
int n, m;int cmp(node a, node b)
{if(a.r == b.r) return a.l > b.l;return a.r < b.r;
}void work()
{LL x;for(int i = 1; i <= n; i++) scanf("%I64d%I64d", &l[i], &r[i]);for(int i = 1; i <= m; i++) {scanf("%I64d", &x);s.insert(mp(x, i));}for(int i = 1; i < n; i++) {p[i].l = l[i+1] - r[i];p[i].r = r[i+1] - l[i];p[i].id = i;}int ok = 1;sort(p+1, p+n, cmp);for(int i = 1; i < n; i++) {it = s.lower_bound(mp(p[i].l, 0));if(it == s.end()) {ok = 0;break;}if(it->first > p[i].r) {ok = 0;break;}ans[p[i].id] = it->second;s.erase(it);}if(!ok) printf("No\n");else {printf("Yes\n");for(int i = 1; i < n; i++) printf("%d%c", ans[i], i == n - 1 ? '\n' : ' ');}}int main()
{while(scanf("%d%d", &n, &m)!=EOF) {work();}return 0;
}

C题：线段树可以做，当然用set做也是可以的，U就找x比他大的， L就找x比他小的。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 200005
#define maxm 2000005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// headset<pii> s;
set<pii>::iterator it;
int x[maxn];
int y[maxn];
char ss[maxn];
int n, m;void work()
{x[0] = y[0] = 0;s.insert(mp(0, m+1));s.insert(mp(n+1, m+1));for(int i = 1; i <= m; i++) {scanf("%d%d%s", &x[i], &y[i], ss);if(ss[0] == 'U') it = s.lower_bound(mp(x[i], -1));else {it = s.upper_bound(mp(x[i], m+1));it--;}if(it->first == x[i]) {printf("0\n");continue;}s.insert(mp(x[i], i));if(ss[0] == 'U') {printf("%d\n", y[i] - y[it->second]);y[i] = y[it->second];}else {printf("%d\n", x[i] - x[it->second]);x[i] = x[it->second];}}
}int main()
{while(scanf("%d%d", &n, &m)!=EOF) {work();}return 0;
}

D题：不是很难的数学题，但是情况有很多。基本思想就是每次找出剩下的线能到达的最远的点，然后每次这样做。找用二分查找，碰到有循环就就取模。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 200005
#define maxm 2000005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// headmap<int, int> mpp;
pii b[maxn];
int a[maxn];
int n, m;void calc(int pos, int x, int dir)
{if(dir == 0 && a[pos] + x < a[pos+1]) {printf("%d\n", b[pos].second);return;}if(dir == 1 && a[pos] - x > a[pos-1]) {printf("%d\n", b[pos].second);return;}if(dir == 0) {int t, nxt;if(a[pos] + x >= a[n]) {t = a[n] - a[pos];nxt = n;}else {nxt = upper_bound(a+1, a+n+1, a[pos]+x) - a - 1;t = a[nxt] - a[pos];}int tt = x / t;if(tt % 2 == 0) calc(pos, x - tt * t, 0);else calc(nxt, x - tt * t, 1);}else {int t, nxt;if(a[pos] - x <= a[1]) {t = a[pos] - a[1];nxt = 1;}else {nxt = lower_bound(a+1, a+n+1, a[pos]-x) - a;t = a[pos] - a[nxt];}int tt = x / t;if(tt % 2 == 0) calc(pos, x - tt * t, 1);else calc(nxt, x - tt * t, 0);}
}void work()
{int t, x;for(int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = mp(a[i], i);sort(b+1, b+n+1);sort(a+1, a+n+1);for(int i = 1; i <= n; i++) mpp[b[i].second] = i;while(m--) {scanf("%d%d", &t, &x);t = mpp[t];if(n == 1) {printf("1\n");continue;}if(a[t] + x >= a[n]) calc(n, x - (a[n] - a[t]), 1);else {int nxt = upper_bound(a+1, a+n+1, a[t]+x) - a - 1;if(nxt == t && nxt == 1) {printf("%d\n", b[1].second);continue;}t = a[nxt] - a[t];calc(nxt, x - t, 1);}}
}int main()
{while(scanf("%d%d", &n, &m)!=EOF) {work();}return 0;
}

E题：先用边双连通缩点，因为对于一个边双连通分量，改变其中的边的方向，一定能把这个双连通变成强连通分量。那么一个边双连通对于这个题就完全符合了。因此我们缩点。缩点以后对于每一个连通块都是一颗树，先预处理出各个树的lca。对于每个s，t，s到t的边的方向一定是确定的。也就是说树上每一个点走出去的方向是确定的。用up，down记录方向，每个询问用标记法处理。最后对于每个树dfs，如果有一个点up和down都有值，那么这是不合法的。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 200005
#define maxm 400005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// headstruct Edge
{int v, next;Edge(int v = 0, int next = 0) : v(v), next(next) {} 
}E[maxm << 1];const int M = 20;stack<int> s;
pii tpair[maxn];
int H[maxn], cntE, h[maxn];
bool used[maxn];
int bcc[maxn];
int block[maxn];
int dep[maxn];
int dfn[maxn];
int vis[maxn];
int low[maxn];
int anc[maxn][M];
int up[maxn];
int down[maxn];
int n, m, q, ok, dfs_clock, T, bcc_cnt;void addedges(int u, int v, int *HH)
{E[cntE] = Edge(v, HH[u]);HH[u] = cntE++;
}void init()
{T = 0;dfs_clock = cntE = bcc_cnt = 0;memset(H, -1, sizeof H);memset(h, -1, sizeof h);memset(vis, 0, sizeof vis);memset(dfn, 0, sizeof dfn);memset(used, 0, sizeof used);
}void tarjan(int u)
{dfn[u] = low[u] = ++dfs_clock;s.push(u), block[u] = T;for(int e = H[u]; ~e; e = E[e].next) if(!used[e / 2]) {used[e / 2] =true;int v = E[e].v;if(!dfn[v]) {tarjan(v);low[u] = min(low[u], low[v]);}else low[u] = min(low[u], dfn[v]);}if(low[u] == dfn[u]) {int t;bcc_cnt++;do {t = s.top();bcc[t] = bcc_cnt;s.pop();}while(t != u);}
}void dfs(int u, int fa)
{vis[u] = true;anc[u][0] = fa;for(int e = h[u]; ~e; e = E[e].next) {int v = E[e].v;if(vis[v]) continue;dep[v] = dep[u] + 1;dfs(v, u);}
}void DFS(int u)
{vis[u] = true;for(int e = h[u]; ~e; e = E[e].next) {int v = E[e].v;if(vis[v]) continue;DFS(v);up[u] += up[v];down[u] += down[v];}if(up[u] && down[u]) ok = 0;
}int to(int u, int d)
{for(int i = M - 1; i >= 0; i--) if(dep[anc[u][i]] >= d) u = anc[u][i];return u;
}int lca(int u, int v)
{if(dep[u] < dep[v]) swap(u, v);u = to(u, dep[v]);for(int i = M - 1; i >= 0; i--) if(anc[u][i] != anc[v][i]) u = anc[u][i], v = anc[v][i];return u == v ? u : anc[u][0];
}void work()
{int u, v;for(int i = 1; i <= m; i++) {scanf("%d%d", &u, &v);tpair[i] = mp(u, v);addedges(u, v, H);addedges(v, u, H); }for(int i = 1; i <= n; i++) if(!dfn[i]) {T++;tarjan(i);}for(int i = 1; i <= m; i++) {u = bcc[tpair[i].first];v = bcc[tpair[i].second];if(u == v) continue;addedges(u, v, h);addedges(v, u, h);}for(int i = 1; i <= bcc_cnt; i++) if(!vis[i]) dfs(i, i);for(int i = 1; i < M; i++)for(int j = 1; j <= n; j++)anc[j][i] = anc[anc[j][i-1]][i-1];ok = 1;for(int i = 1; i <= q; i++) {scanf("%d%d", &u, &v);if(block[u] != block[v]) ok = 0;u = bcc[u];v = bcc[v];int c = lca(u, v);up[u]++;up[c]--;down[v]++;down[c]--;}memset(vis, 0, sizeof vis);for(int i = 1; i <= bcc_cnt; i++) if(!vis[i]) DFS(i);printf("%s\n", ok ? "Yes" : "No");
}int main()
{while(scanf("%d%d%d", &n, &m, &q)!=EOF) {init();work();}return 0;
}