标签：enum const ons mes include struct false 图片 cli
OJ 1190
二分答案，对每个答案进行判断，用并查集进行优化，将整个场地的左侧和右侧分别设为0和n+1，每个点覆盖的范围设为i，$O(n^2)$判断点之间是否连通，最后只需判断0和n+1是否连通即可。
#include <iostream> #include <cstdio> #include <cstring> using namespace std; inline long long read() { long long ans = 0, f = 1; char ch = getchar(); while(ch < ‘0‘ || ch > ‘9‘) f *= (ch == ‘-‘) ? -1 : 1, ch = getchar(); do ans = ((ans << 1) + (ans << 3) + (ch ^ 48)), ch = getchar(); while(ch >= ‘0‘ && ch <= ‘9‘); return ans * f; } const int MAXN = 2001; const double esp = 1e-7; double a[MAXN], b[MAXN], x, y; int fa[MAXN], n; inline int get(int x){ return (fa[x] == x) ? x : fa[x] = get(fa[x]); } bool judge(double val){ for(int i=0; i<=n+1; i++) fa[i] = i; for(int i=1; i<=n; i++){ if(a[i] < val + esp || b[i] + val + esp > y) fa[get(i)] = get(0); if(b[i] < val + esp || a[i] + val + esp > x) fa[get(i)] = get(n+1); } for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) if(i != j && (a[j] - a[i]) * (a[j] - a[i]) + (b[j] - b[i]) * (b[j] - b[i]) < val * val * 4 + esp) fa[get(i)] = get(j); return get(0) != get(n+1); } int main(){ x = read(), y = read(), n = read(); for(int i=1; i<=n; i++) a[i] = read(), b[i] = read(); double l = 0, r = max(x, y); while(l + 1e-3 < r){ double mid = (l + r) / 2; if(judge(mid)) l = mid; else r = mid; } printf("%.2lf", l); return 0; }
OJ 1191
整个图是一个基环树，要求基环树直径，分为两种情况，一个是路径经过环，一个路径不经过环，第二种情况直接Dp求直径，对第一种情况使用动态规划
$ans = max(ans, Dp[i]+Dp[j]+sum[i]-sum[j]),i>j-len,sum[i]-sum[j]<=sum[len]/2$，单调队列优化即可
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(ch < ‘0‘ || ch > ‘9‘) f *= (ch == ‘-‘) ? -1 : 1, ch = getchar(); do ans = ((ans << 1) + (ans << 3) + (ch ^ 48)), ch = getchar(); while(ch >= ‘0‘ && ch <= ‘9‘); return ans * f; } const int MAXN = 1000005; int head[MAXN], next[MAXN<<1], last[MAXN<<1], val[MAXN<<1], lineNum; void add(int x,int y,int v){ lineNum++, next[lineNum] = head[x], last[lineNum] = y, val[lineNum] = v, head[x] = lineNum; } int cir[MAXN<<1], top, len, sta[MAXN], vis[MAXN]; long long ans, ret, Dp[MAXN], ext[MAXN<<1]; int deq[MAXN<<1], L = 0, R = -1; struct Edge{ int x, y, val; const bool operator < (const Edge &temp) const{ if(x != temp.x) return x < temp.x; if(y != temp.y) return y < temp.y; return val < temp.val; } }e[MAXN]; void findCir(int x,int fa){ sta[++top] = x, vis[x] = 1; for(int l = head[x]; l; l = next[l]){ int y = last[l]; if(y == fa) continue; if(vis[y] == 1){ do cir[++len] = sta[top], vis[sta[top]] = 2; while(sta[top--] != y); continue; } if(!vis[y]) findCir(y, x); } if(vis[x] == 1) vis[x] = 2, top--; } void dfs(int x,int fa){ for(int l = head[x]; l; l = next[l]){ int y = last[l]; if(y == fa || vis[y]) continue; dfs(y, x); ret = max(ret, Dp[x] + Dp[y] + val[l]); Dp[x] = max(Dp[x], Dp[y] + val[l]); } } int main(){ int n = read(); for(int i=1; i<=n; i++){ e[i].x = read(), e[i].y = read(), e[i].val = read(); if(e[i].x > e[i].y) swap(e[i].x, e[i].y); } sort(e+1, e+1+n); bool flag = false; for(int i=1; i<=n; i++){ if(e[i].x == e[i].y) flag = true; else if(e[i].x == e[i-1].x && e[i].y == e[i-1].y) flag = true; else add(e[i].x, e[i].y, e[i].val), add(e[i].y, e[i].x, e[i].val); } if(flag){ dfs(1, 0); printf("%lld", ret + 1); return 0; } findCir(1, -1); cir[0] = cir[len]; for(int i=len+1; i<=len*2; i++) cir[i] = cir[i-len]; for(int i=1; i<=len; i++) for(int l = head[cir[i]]; l; l = next[l]) if(last[l] == cir[i-1]) ext[i] = val[l]; for(int i=len+1; i<=len*2; i++) ext[i] = ext[i-len]; for(int i=1; i<=len*2; i++) ext[i] += ext[i-1]; memset(vis, 0, sizeof(vis)); for(int i=1; i<=len; i++){ vis[cir[i-1]] = vis[cir[i+1]] = true; ret = 0, dfs(cir[i], -1); vis[cir[i-1]] = vis[cir[i+1]] = false; ans = max(ans, ret); } for(int i=1; i<=len*2; i++){ while(L <= R && (deq[L] < i - len || ext[i] - ext[deq[L]] > ext[len] / 2)) L++; if(L <= R) ans = max(ans, Dp[cir[i]] + Dp[cir[deq[L]]] + ext[i] - ext[deq[L]]); while(L <= R && Dp[cir[deq[R]]] - ext[deq[R]] <= Dp[cir[i]] - ext[i]) R--; deq[++R] = i; } printf("%lld", ans + 1); return 0; }
OJ 1192
使用树形Dp，Dp[MAXN][2]，第二维表示以x为根的子树使用（不使用）传送阵的最短时间，转移方程：
$Dp[x][0] += Dp[y][0] + 2*val[l],Dp[x][1] += min(Dp[y][1] + 2*val[l], Dp[y][0] + val[l] - d[y])$
#include <iostream> #include <cstdio> #include <cstring> using namespace std; long long read(){ long long ans = 0, f = 1; char ch = getchar(); while(ch < ‘0‘ || ch > ‘9‘) f *= (ch == ‘-‘) ? -1 : 1, ch = getchar(); do ans = ((ans << 1) + (ans << 3) + (ch ^ 48)), ch = getchar(); while(ch >= ‘0‘ && ch <= ‘9‘); return ans * f; } const int MAXN = 1000001; int head[MAXN], next[MAXN<<1], last[MAXN<<1], val[MAXN<<1], lineNum; void add(int x,int y,int v){ lineNum++, next[lineNum] = head[x], last[lineNum] = y, val[lineNum] = v, head[x] = lineNum; } long long Dp[MAXN][2], d[MAXN]; void dfs(int x,int fa){ for(int l = head[x]; l; l = next[l]){ int y = last[l]; if(y == fa) continue; dfs(y, x); d[x] = max(d[x], d[y] + val[l]); Dp[x][0] += Dp[y][0] + val[l] * 2; Dp[x][1] += min(Dp[y][1] + val[l] * 2, Dp[y][0] + val[l] - d[y]); } } int main(){ int n = read(); for(int i=1; i<n; i++){ int x = read(), y = read(), v = read(); add(x, y, v), add(y, x, v); } dfs(1, -1); printf("%lld", Dp[1][1]); return 0; }
标签：enum const ons mes include struct false 图片 cli
原文地址：https://www.cnblogs.com/PHDHD/p/12383429.html
