Description
小明喜欢看直播，他订阅了很多主播，主播们有固定的直播时间 [Li, Ri] 。
可是他网速只有2M，不能同时播放两个直播，所以同一时间只能看一个直播。
并且他只会去看能完整看完的直播（从开播到停播都能观看）。
他想知道最多能看多长时间的直播呢？
注意 [1, 3] 和 [3, 4] 不能两者都选择。
Input
第一行一个N。
接下来N行每行两个整数Li, Ri。
1 <= N <= 2e5
1 <= Li <= Ri <= 1e9
Output
输出最多能看的时间。
Sample Input 1
3 1 3 2 6 5 8
Sample Output 1
7
这个题目不是特别难，可以用很多方法写。
#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <map> #include <queue> #include <vector> #define inf 0x3f3f3f3f using namespace std; typedef long long ll; const int INF = 0x3f3f3f3f; const int maxn = 4e5 + 10; struct node { ll l, r, com; node(ll l=0,ll r=0,ll com=0):l(l),r(r),com(com){} }ex[maxn]; ll dp[maxn], a[maxn]; struct edge { int ed, w; edge(int ed=0,int w=0):ed(ed),w(w){} }; vector<edge>e[maxn]; int main() { int n, tot = 0; scanf("%d", &n); for(int i=1;i<=n;i++) { int l, r; scanf("%d%d", &l, &r); ex[i] = node(l, r, 0); a[++tot] = l; a[++tot] = r; } sort(a + 1, a + 1 + tot); int len = unique(a + 1, a + 1 + tot) - a - 1; for(int i=1;i<=n;i++) { ex[i].com = ex[i].r - ex[i].l + 1; ex[i].l = lower_bound(a + 1, a + 1 + len, ex[i].l) - a; ex[i].r = lower_bound(a + 1, a + 1 + len, ex[i].r) - a; e[ex[i].l].push_back(edge(ex[i].r,ex[i].com)); } for(int i=len;i>=1;i--) { dp[i] = dp[i + 1]; for(int j=0;j<e[i].size();j++) { edge x = e[i][j]; dp[i] = max(dp[i], dp[x.ed + 1] + x.w); } } printf("%lld\n", dp[1]); return 0; }
Dragon plus 14:50:56 #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #define fir first #define sec second using namespace std; typedef pair<int,int> P; const int maxn = 2e5+7; int n,len; P s[maxn]; int v[maxn<<2]; int Max[maxn<<4]; bool cmp(const P& a,const P& b) { return a.sec<b.sec; } void pushup(int num) {Max[num] = max(Max[num<<1],Max[num<<1|1]);} void build(int l,int r,int num) { if(l==r) { Max[num]=0; return ; } int mid = (l+r) >>1; build(l,mid,num<<1); build(mid+1,r,num<<1|1); pushup(num); } void modify(int l,int r,int num,int pos,int mods) { if(pos<l || r <pos) return ; if(pos<=l&&r<=pos) { Max[num] = max(Max[num],mods); return ; } int mid = (l+r) >>1; if(pos<=mid) modify(l,mid,num<<1,pos,mods); if(mid< pos) modify(mid+1,r,num<<1|1,pos,mods); pushup(num); } int quriy(int l,int r,int num,int le,int ri) { if(ri<l || r<le) return 0; if(le<=l && r<=ri) return Max[num]; int mid = (l+r) >>1; int ans = 0; if(le<=mid) ans = max(ans,quriy(l,mid,num<<1,le,ri)); if(mid< ri) ans = max(ans,quriy(mid+1,r,num<<1|1,le,ri)); return ans; } int main() { int now =1; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d%d",&s[i].fir,&s[i].sec); v[now++] = s[i].fir; v[now++] = s[i].sec; } sort(v,v+now); sort(s,s+n,cmp); len = unique(v,v+now)-v; build(1,len,1); for(int i=0;i<n;i++) { int res = s[i].sec -s[i].fir+1; int d = lower_bound(v,v+len,s[i].fir)-v; int d2 = lower_bound(v,v+len,s[i].sec)-v+1; res += quriy(1,len,1,1,d); modify(1,len,1,d2,res); } printf("%d\n",quriy(1,len,1,1,len)); return 0; }
#include <iostream> #include <algorithm> #include <cstdio> #define fir first #define sec second using namespace std; typedef pair<int,int> P; const int maxn = 2e5+7; bool cmp(const P&a,const P&b) { return a.sec<b.sec; } int n,lst; P s[maxn]; int dp[maxn<<2]; int v[maxn<<2]; int p[maxn]; int main() { int now = 1; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d%d",&s[i].fir,&s[i].sec); v[now++]=s[i].fir; v[now++]=s[i].sec; } int ans = 0; sort(v,v+now); sort(s,s+n,cmp); int len = unique(v,v+now)-v; for(int i=0;i<n;i++) { int d = s[i].sec - s[i].fir+1; int l = lower_bound(v,v+len,s[i].fir)-v; int r = lower_bound(v,v+len,s[i].sec)-v; int Max = lower_bound(p,p+lst+1,l)-p-1; //cout<<p[Max]<<' '<<dp[p[Max]]<<endl; dp[r] = max(dp[r],dp[p[Max]]+d); ans = max(dp[r],ans); //cout<<r<<' '<<dp[r]<<' '<<lst<<endl; if(dp[p[lst]] >= dp[r]) continue; else { if(p[lst] != r) p[++lst] = r; } } printf("%d\n",ans ); return 0; } CSUST Online Judge Powered by OnlineJudge Version: 20190727-ed4d6
#include <bits/stdc++.h> using namespace std; const int MAX = 2e5 + 50; struct date { int le, ri; } a[MAX]; int lsh[MAX * 2]; bool cmp(date A, date B) { if (A.le != B.le) { return A.le < B.le; } else { return A.ri < B.ri; } } int dp[MAX * 2]; int s[MAX * 2]; int main() { int n; scanf("%d", &n); int cnt = 0; for (int i = 1; i <= n; i++) { scanf("%d%d", &a[i].le, &a[i].ri); lsh[++cnt] = a[i].le; lsh[++cnt] = a[i].ri; } sort(lsh + 1, lsh + cnt + 1); cnt = unique(lsh + 1, lsh + cnt + 1) - lsh - 1; sort(a + 1, a + n + 1, cmp); for (int i = 1; i <= n; i++) { a[i].le = lower_bound(lsh + 1, lsh + cnt + 1, a[i].le) - lsh; a[i].ri = lower_bound(lsh + 1, lsh + cnt + 1, a[i].ri) - lsh; } int k = 1; int top = 1, tail = 0; for (int i = 1; i <= cnt && k <= n; i++) { while (i == a[k].le) { dp[a[k].ri] = max(dp[a[k].ri], lsh[a[k].ri] - lsh[a[k].le] + s[top] + 1); k++; } while (top <= tail && s[tail] <= dp[i]) { tail--; } s[++tail] = dp[i]; } int ans = 0; for (int i = 1; i <= cnt; i++) { ans = max(ans, dp[i]); } printf("%d\n", ans); return 0; }