标签：txt 设置 初始 种类 printf sizeof 复杂 ast getchar
DP&图论 DAY 6 下午 考试
3 5 10 3 1 3 437 1 2 282 1 5 328 1 2 519 1 2 990 2 3 837 2 4 267 2 3 502 3 5 613 4 5 132 1 3 4 10 13 4 1 6 484 1 3 342 2 3 695 2 3 791 2 8 974 3 9 526 4 9 584 4 7 550 5 9 914 6 7 444 6 8 779 6 10 350 8 8 394 9 10 3 7 10 9 4 1 2 330 1 3 374 1 6 194 2 4 395 2 5 970 2 10 117 3 8 209 4 9 253 5 7 864 8 5 10 6
437 526 641
题解
>50 pt dij 跑暴力
（Floyd太慢了QWQ O（n^3））
枚举每个点作为起点，dijkstra，跑暴力 O( (n+m)logn )，寻找全局最短路
#include<iostream> #include<cstdio> #include<string> #include<cstring> #include<cmath> #include<cstdlib> #include<algorithm> #include<queue> using namespace std; inline int read() { int ans=0; char last=‘ ‘,ch=getchar(); while(ch<‘0‘||ch>‘9‘) last=ch,ch=getchar(); while(ch>=‘0‘&&ch<=‘9‘) ans=ans*10+ch-‘0‘,ch=getchar(); if(last==‘-‘) ans=-ans; return ans; } typedef pair<int,int> pa; const int maxn=1e5+10; const int maxm=2e5+10; int T; int n,m,k; struct node { int to,nxt,dis; }edge[maxm]; int head[maxn],cnt=0; void addedge(int u,int v,int w) { edge[++cnt].to =v;edge[cnt].dis =w;edge[cnt].nxt =head[u];head[u]=cnt; edge[++cnt].to =u;edge[cnt].dis =w;edge[cnt].nxt =head[v];head[v]=cnt; } int dis[maxn]; int a[maxn]; void dijkstra(int s) { priority_queue<pa,vector<pa>,greater<pa> >q; q.push(make_pair(s,dis[s]=0)); while(!q.empty() ) { pa now=q.top(); q.pop() ; if(now.second !=dis[now.first ]) continue; for(int i=head[now.first ],v;i;i=edge[i].nxt ) { if(v=edge[i].to ,dis[v]>dis[now.first ]+edge[i].dis ) q.push(make_pair(v,dis[v]=dis[now.first]+edge[i].dis)); } } } int main() { T=read(); for(int t=1;t<=T;t++) { cnt=0; memset(head,0,sizeof(head)); memset(edge,0,sizeof(edge)); memset(a,0,sizeof(a)); n=read();m=read();k=read(); for(int i=1;i<=m;i++) { int u=read(),v=read(),w=read(); addedge(u,v,w); } for(int i=1;i<=k;i++) a[i]=read(); int ans=0x3f3f3f3f; for(int i=1;i<=k;i++) { memset(dis,0x3f,sizeof(dis)); dijkstra(a[i]); for(int j=1;j<=k;j++) { if(a[i]==a[j]) continue; ans=min(ans,dis[a[j]]); } } if(ans>=0x3f3f3f3f) ans=-1; printf("%d\n",ans); } return 0; }
>100pt 考虑优化枚举量
因为答案是两个编号不同的点，所以对应的二进制编码至少有一位不同
枚举二进制的每一位
假设枚举到第 i 位，把这一位是 1 的点设置为源点，是 0 的设置为汇点，跑一遍多源多汇最短路
设置一个超级源点，向所有第一层集合的点连一条长度为 0 的边
设置一个超级汇点，所有最后一个集合的点向超级汇点连一条长度为 0 的边
跑从超级源点到超级汇点的最短路
跑最多32次就可以得到答案
这两个集合既可以是 1~n ，也可以是 1~k
显然 1~k 更优
#include <queue> #include <cstdio> #include <cstring> template <class cls> inline cls min(const cls & a, const cls & b) { return a < b ? a : b; } const int mxn = 100005; const int mxm = 500005; const int inf = 0x3f3f3f3f; int n, m, k; int points[mxn]; int tot; int hd[mxn]; int nt[mxm]; int to[mxm]; int vl[mxm]; inline void add_edge(int u, int v, int w) { nt[++tot] = hd[u]; to[tot] = v; vl[tot] = w; hd[u] = tot; } int dis[mxn]; struct data { int u, d; data(int _u, int _d) : u(_u), d(_d) {} bool operator < (const data & that) const { return d > that.d; } }; std::priority_queue<data> heap; int main() { int cas; scanf("%d", &cas); for (int c = 0; c < cas; ++c) { scanf("%d%d%d", &n, &m, &k); memset(hd, 0, sizeof(int) * (n + 5)); tot = 0; for (int i = 0, u, v, w; i < m; ++i) { scanf("%d%d%d", &u, &v, &w); add_edge(u, v, w); add_edge(v, u, w); } for (int i = 0; i < k; ++i) scanf("%d", points + i); int ans = inf; for (int i = 1; i < k; i <<= 1) { memset(dis, inf, sizeof(int) * (n + 5)); for (int j = 0, p; j < k; ++j) if (p = points[j], (j & i) == 0) heap.push(data(p, dis[p] = 0)); while (!heap.empty()) { int u = heap.top().u; int d = heap.top().d; heap.pop(); if (dis[u] != d) continue; for (int e = hd[u], v, w; e; e = nt[e]) if (v = to[e], w = vl[e], dis[v] > d + w) heap.push(data(v, dis[v] = d + w)); } for (int j = 0, p; j < k; ++j) if (p = points[j], (j & i) != 0) ans = min(ans, dis[p]); } printf("%d\n", ans == inf ? -1 : ans); } return 0; }
3 5 10 5 4 10 8 1 10 1 3 1 4 1 5 1 3 2 1 2 5 4 3 4 3 4 5 5 1 1 4 4 6 1 9 4 7 2 9 5 10 5 2 8 8 10 10 2 1 2 3 3 2 3 4 3 1 3 2 3 4 4 1 5 4 5 1 1 4 2 3 4 7 3 10 1 5 5 10 5 9 9 8 2 1 1 5 1 5 2 1 2 4 2 4 2 4 3 2 3 1 4 3 4 3 5 9 3 9 2 7 5 1 5 4
40 60 90 70 90 8 30 70 100 10 9 81 63 14
题解
50pt dfs 暴力
观察题目发现我们只需要找到对于一个点的 
就好
#include<iostream> #include<cstdio> #include<string> #include<cstring> #include<cmath> #include<cstdlib> #include<algorithm> #include<queue> using namespace std; typedef long long ll; inline ll read() { ll ans=0; char last=‘ ‘,ch=getchar(); while(ch<‘0‘||ch>‘9‘) last=ch,ch=getchar(); while(ch>=‘0‘&&ch<=‘9‘) ans=ans*10+ch-‘0‘,ch=getchar(); if(last==‘-‘) ans=-ans; return ans; } const ll maxn=2e5+10; const ll maxm=4e5+10; ll T; ll n,m,k; ll w[maxn],son[maxn]; ll head[maxn],to[maxm],nxt[maxn]; bool vis[maxn]; void addedge(ll u,ll v,ll cnt) { to[cnt]=v;nxt[cnt]=head[u];head[u]=cnt; } void chuli(ll u) { if(head[u]==0) return; if(vis[u]) return; vis[u]=1; for(ll i=head[u];i;i=nxt[i]) { ll v=to[i]; chuli(v); if(w[son[v]]>w[son[u]]) son[u]=son[v]; } return; } int main() { // freopen("neural.in","r",stdin); // freopen("ceshi.txt","w",stdout); T=read(); for(ll t=1;t<=T;t++) { memset(w,0,sizeof(w)); memset(son,0,sizeof(son)); memset(head,0,sizeof(head)); memset(nxt,0,sizeof(nxt)); memset(to,0,sizeof(to)); memset(vis,0,sizeof(vis)); n=read();m=read();k=read(); for(ll i=1;i<=n;i++) w[i]=read(),son[i]=i; for(ll i=1;i<=m;i++) { ll u,v; u=read();v=read(); addedge(u,v,i); } for(ll i=1;i<=n;i++) chuli(i); for(ll i=1;i<=k;i++) { ll u,x; u=read();x=read(); printf("%lld\n",(long long)x*w[son[u]]); } } return 0; }
>100pt
建反向边，tarjan然后拓扑就行了
( 然后我们发现一个大佬ych的思路
思路是tarjan缩点，一个强连通分量的初始ans就是这个强连通分量里面点的最大值。然后建立新图，找到入度为0的点开始dfs，然后更新强连通分量的ans。
询问点就是找点所在的强连通分量，输出强连通分量的ans就ok )
#include <cstdio> #include <cstring> template <class cls> inline cls min(const cls & a, const cls & b) { return a < b ? a : b; } template <class cls> inline cls max(const cls & a, const cls & b) { return a > b ? a : b; } const int mxn = 200005; const int mxm = 400005; int n, m, k, w[mxn]; struct edge { int u, v; } edges[mxm]; int tot; int hd[mxn]; int to[mxm << 1]; int nt[mxm << 1]; inline void add_edge(int u, int v) { nt[++tot] = hd[u]; to[tot] = v; hd[u] = tot; } int tim; int cnt; int top; int dfn[mxn]; int low[mxn]; int stk[mxn]; int scc[mxn]; void tarjan(int u) { dfn[u] = low[u] = ++tim; stk[++top] = u; for (int e = hd[u], v; e; e = nt[e]) if (v = to[e], scc[v] == 0) { if (dfn[v] == 0)tarjan(v), low[u] = min(low[u], low[v]); else low[u] = min(low[u], dfn[v]); } if (dfn[u] == low[u]) { cnt += 1; do { scc[stk[top]] = cnt; } while (stk[top--] != u); } } int oe[mxn]; int mx[mxn]; int que[mxn]; void bfs() { int l = 0, r = 0; for (int i = 1; i <= cnt; ++i) if (oe[i] == 0) que[r++] = i; while (l < r) { int u = que[l++]; for (int e = hd[u], v; e; e = nt[e]) if (v = to[e], mx[v] = max(mx[v], mx[u]), --oe[v] == 0) que[r++] = v; } } int main() { int cas; scanf("%d", &cas); for (int c = 0; c < cas; ++c) { scanf("%d%d%d", &n, &m, &k); for (int i = 1; i <= n; ++i) scanf("%d", w + i); memset(hd, 0, sizeof(int) * (n + 5)); tot = 0; for (int i = 0; i < m; ++i) { scanf("%d%d", &edges[i].u, &edges[i].v); add_edge(edges[i].u, edges[i].v); } tim = cnt = top = 0; memset(scc, 0, sizeof(int) * (n + 5)); memset(dfn, 0, sizeof(int) * (n + 5)); for (int i = 1; i <= n; ++i) if (scc[i] == 0) tarjan(i); memset(hd, 0, sizeof(int) * (cnt + 5)); tot = 0; memset(oe, 0, sizeof(int) * (cnt + 5)); memset(mx, 0, sizeof(int) * (cnt + 5)); for (int i = 0; i < m; ++i) { int u = scc[edges[i].u]; int v = scc[edges[i].v]; if (u != v) add_edge(v, u), oe[u] += 1; } for (int i = 1; i <= n; ++i) mx[scc[i]] = max(mx[scc[i]], w[i]); bfs(); for (int i = 0, u, x; i < k; ++i) { scanf("%d%d", &u, &x); printf("%lld\n", 1LL * x * mx[scc[u]]); } } return 0; }
样例输入
2 3 1 0 2 0 2 2 1 0
题解
很像 Qtree 啊 所以一样hintai
树链剖分
单点修改，查询区间内值为x的数
考虑如何实现？？？
如果x比较少，完全可以建几棵线段树来实现，就好比 20% 的数据，颜色种类不超过 5
每次修改一个颜色，就是在该颜色线段树内 +1，原颜色线段树内 -1
颜色种类多了怎么办？
暴力：开100个树状数组，和刚才没什么区别
如果线段树在每一个节点上维护一个100的数组
合并的时候可以直接暴力统计节点次数，这样代价是区间长度
如果每一位枚举则是n*100
每一层访问的点是n的，一共log层
复杂度 O（nlogn）
继续优化：
离线操作，只需要建一棵线段树
操作分类，与同一种颜色有关的操作放到一起
所有操作次数相加就是2m
所以操作还是o(m)
代码
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; inline int getint() { int r = 0, c = getchar(); for (; c < 48; c = getchar()); for (; c > 47; c = getchar()) r = r * 10 + c - 48; return r; } const int mxc = 100005; const int mxn = 100005; const int mxm = 200005; int n, m, c; int tt; int hd[mxn]; int to[mxm]; int nt[mxm]; inline void addedge(int x, int y) { nt[++tt] = hd[x], to[tt] = y, hd[x] = tt; nt[++tt] = hd[y], to[tt] = x, hd[y] = tt; } struct data { int k, x, y; data() {} ; data(int a, int b, int c) : k(a), x(b), y(c) {} ; }; int color[mxn]; #include <vector> vector<data> vec[mxc]; int tim; int dfn[mxn]; int top[mxn]; int fat[mxn]; int dep[mxn]; int son[mxn]; int siz[mxn]; void dfs1(int u, int f) { siz[u] = 1; son[u] = 0; fat[u] = f; dep[u] = dep[f] + 1; for (int i = hd[u], v; i; i = nt[i]) if (v = to[i], v != f) { dfs1(v, u); siz[u] += siz[v]; if (siz[v] > siz[son[u]]) son[u] = v; } } void dfs2(int u, int f) { dfn[u] = ++tim; if (son[f] == u) top[u] = top[f]; else top[u] = u; if (son[u]) dfs2(son[u], u); for (int i = hd[u], v; i; i = nt[i]) if (v = to[i], v != f && v != son[u]) dfs2(v, u); } int bit[mxn]; inline void add(int p, int v) { for (; p <= n; p += p & -p) bit[p] += v; } inline int ask(int l, int r) { int sum = 0; --l; for (; r; r -= r & -r) sum += bit[r]; for (; l; l -= l & -l) sum -= bit[l]; return sum; } int ans[mxn]; signed main() { int cas = getint(); while (cas--) { n = getint(); m = getint(); for (int i = 1; i <= n; ++i) vec[color[i] = getint()].push_back(data(0, i, +1)); c = 0; for (int i = 1; i <= n; ++i) c = max(c, color[i]); memset(hd, 0, sizeof(int) * (n + 5)); tt = 0; for (int i = 1; i < n; ++i) { int x = getint(); int y = getint(); addedge(x, y); } for (int i = 1; i <= m; ++i) { if (getint() == 1) { int p = getint(); int a = color[p]; int b = color[p] = getint(); vec[a].push_back(data(0, p, -1)); vec[b].push_back(data(0, p, +1)); } else { int x = getint(); int y = getint(); int k = getint(); vec[k].push_back(data(i, x, y)); } } dfs1(1, 0); dfs2(1, 0); memset(ans, -1, sizeof ans); for (int k = 1; k <= c; ++k) { int sz = vec[k].size(); memset(bit, 0, sizeof bit); for (int i = 0; i < sz; ++i) { const data &d = vec[k][i]; ans[d.k] = 0; if (d.k == 0) add(dfn[d.x], d.y); else { int a = d.x, ta = top[a]; int b = d.y, tb = top[b]; while (ta != tb) { if (dep[ta] >= dep[tb]) ans[d.k] += ask(dfn[ta], dfn[a]), ta = top[a = fat[ta]]; else ans[d.k] += ask(dfn[tb], dfn[b]), tb = top[b = fat[tb]]; } if (dep[a] <= dep[b]) ans[d.k] += ask(dfn[a], dfn[b]); else ans[d.k] += ask(dfn[b], dfn[a]); } } } for (int i = 1; i <= m; ++i) if (ans[i] >= 0) printf("%d\n", ans[i]); for (int i = 1; i <= c; ++i) vec[i].clear(); tim = 0; } return 0; }
标签：txt 设置 初始 种类 printf sizeof 复杂 ast getchar
原文地址：https://www.cnblogs.com/xiaoyezi-wink/p/11336877.html
