http://www.lydsy.com/JudgeOnline/problem.php?id=1672

dp很好想，但是是n^2的。。但是可以水過(guò)。。（5s啊。。）

按左端點(diǎn)排序后

f[i]表示取第i個(gè)的最小費(fèi)用

f[i]=min(f[j])+w[i] 當(dāng)j的結(jié)束時(shí)間>=i的開(kāi)始時(shí)間-1

答案就是所有的i滿足i的結(jié)束時(shí)間>=結(jié)束時(shí)間-1

#include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> using namespace std; #define rep(i, n) for(int i=0; i<(n); ++i) #define for1(i,a,n) for(int i=(a);i<=(n);++i) #define for2(i,a,n) for(int i=(a);i<(n);++i) #define for3(i,a,n) for(int i=(a);i>=(n);--i) #define for4(i,a,n) for(int i=(a);i>(n);--i) #define CC(i,a) memset(i,a,sizeof(i)) #define read(a) a=getint() #define print(a) printf("%d", a) #define dbg(x) cout << #x << " = " << x << endl #define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; } inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; } inline const int max(const int &a, const int &b) { return a>b?a:b; } inline const int min(const int &a, const int &b) { return a<b?a:b; }const int N=10005; int f[N], n, ans=~0u>>1; struct dat { int x, y, w; }a[N]; const bool cmp(const dat &x, const dat &y) { return x.x==y.x ? x.y<y.y : x.x<y.x; } int main() {read(n); a[0].y=getint()-1; a[n+1].x=getint()+1;for1(i, 1, n) read(a[i].x), read(a[i].y), read(a[i].w);sort(a+1, a+1+n, cmp);CC(f, 0x3f);f[0]=0;for1(i, 1, n+1) {int s=a[i].x, w=a[i].w;rep(j, i) if(s-1<=a[j].y) f[i]=min(f[i], f[j]+w);}for1(i, 1, n) if(a[i].y>=a[n+1].x-1 && ans>f[i]) ans=f[i];ans=min(ans, f[n+1]);if(ans==f[n+2]) ans=-1;print(ans);return 0; }

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線段樹(shù)：

（調(diào)試了n久啊。。果然不能看別人代碼對(duì)著抄。。。。。）

對(duì)于一個(gè)點(diǎn)i，我們更新它時(shí)要用它所在的區(qū)間（開(kāi)始時(shí)間-1到結(jié)束時(shí)間）有沒(méi)有已經(jīng)接上的（即更新過(guò)的）。

所以我們按左端點(diǎn)排序后

依次更新線段樹(shù)。（單點(diǎn)更新即可，按結(jié)束時(shí)間更新， 首先要詢問(wèn)本區(qū)間，然后接上）

然后詢問(wèn)的時(shí)候直接詢問(wèn)本區(qū)間的答案。

最后答案就是end的單點(diǎn)詢問(wèn)

#include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> using namespace std; #define rep(i, n) for(int i=0; i<(n); ++i) #define for1(i,a,n) for(int i=(a);i<=(n);++i) #define for2(i,a,n) for(int i=(a);i<(n);++i) #define for3(i,a,n) for(int i=(a);i>=(n);--i) #define for4(i,a,n) for(int i=(a);i>(n);--i) #define CC(i,a) memset(i,a,sizeof(i)) #define read(a) a=getint() #define print(a) printf("%d", a) #define dbg(x) cout << #x << " = " << x << endl #define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; } inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; } inline const int max(const int &a, const int &b) { return a>b?a:b; } inline const int min(const int &a, const int &b) { return a<b?a:b; } #define lc x<<1 #define rc x<<1|1 #define MID (l+r)>>1 #define lson l, m, lc #define rson m+1, r, rcconst int N=10005, oo=~0u>>1; int n, st, en; int ans; struct dat { int x, y, w; }a[N]; struct node { int mn; }t[500005]; const bool cmp(const dat &x, const dat &y) { return x.x<y.x; } void pushup(int x) { t[x].mn=min(t[lc].mn, t[rc].mn); } void build(int l, int r, int x) {t[x].mn=oo;if(l==r) return;int m=MID;build(lson); build(rson); } void update(int l, int r, int x, const int &key, const int &R) {if(l==r) {t[x].mn=min(key, t[x].mn);return;}int m=MID;if(R<=m) update(lson, key, R); else update(rson, key, R);pushup(x); } int query(int l, int r, int x, const int &L, const int &R) {if(L<=l && r<=R) return t[x].mn;int m=MID; int mn=oo;if(L<=m) mn=min(mn, query(lson, L, R)); if(m<R) mn=min(mn, query(rson, L, R));return mn; } int main() {read(n); read(st); read(en);for1(i, 1, n) read(a[i].x), read(a[i].y), read(a[i].w);sort(a+1, a+1+n, cmp);build(st, en, 1);for1(i, 1, n) {if(a[i].x<=st) ans=0; else ans=query(st, en, 1, a[i].x-1, a[i].y);if(ans!=oo) update(st, en, 1, ans+a[i].w, a[i].y);}ans=query(st, en, 1, en, en);if(ans==oo) puts("-1");else printf("%d", ans);return 0; }

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Description

Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most obliging of farmers, has no choice but hire some of the cows to clean the barn. Farmer John has N (1 <= N <= 10,000) cows who are willing to do some cleaning. Because dust falls continuously, the cows require that the farm be continuously cleaned during the workday, which runs from second number M to second number E during the day (0 <= M <= E <= 86,399). Note that the total number of seconds during which cleaning is to take place is E-M+1. During any given second M..E, at least one cow must be cleaning. Each cow has submitted a job application indicating her willingness to work during a certain interval T1..T2 (where M <= T1 <= T2 <= E) for a certain salary of S (where 0 <= S <= 500,000). Note that a cow who indicated the interval 10..20 would work for 11 seconds, not 10. Farmer John must either accept or reject each individual application; he may NOT ask a cow to work only a fraction of the time it indicated and receive a corresponding fraction of the salary. Find a schedule in which every second of the workday is covered by at least one cow and which minimizes the total salary that goes to the cows.

約翰的奶牛們從小嬌生慣養(yǎng)，她們無(wú)法容忍牛棚里的任何臟東西．約翰 發(fā)現(xiàn)，如果要使這群有潔癖的奶牛滿意，他不得不雇傭她們中的一些來(lái)清掃牛棚,?約翰的奶牛中有N(1≤N≤10000)頭愿意通過(guò)清掃牛棚來(lái)掙一些零花 錢(qián)．由于在某個(gè)時(shí)段中奶牛們會(huì)在牛棚里隨時(shí)隨地地亂扔垃圾，自然地，她們要求在這段時(shí)間里，無(wú)論什么時(shí)候至少要有一頭奶牛正在打掃．需要打掃的時(shí)段從某一 天的第M秒開(kāi)始，到第E秒結(jié)束f0≤M≤E≤86399)．注意這里的秒是指時(shí)間段而不是時(shí)間點(diǎn)，也就是說(shuō)，每天需要打掃的總時(shí)間是E-M+I秒．?約翰已經(jīng)從每頭牛那里得到了她們?cè)敢饨邮艿墓ぷ饔?jì)劃：對(duì)于某一頭牛，她每天都愿意在笫 Ti，.T2秒的時(shí)間段內(nèi)工作(M≤Ti≤馬≤E)，所要求的報(bào)酬是S美元(0≤S≤500000)．與需打掃時(shí)段的描述一樣，如果一頭奶牛愿意工作的時(shí) 段是每天的第10_20秒，那她總共工作的時(shí)間是11秒，而不是10秒．約翰一旦決定雇傭某一頭奶牛，就必須付給她全額的工資，而不能只讓她工作一段時(shí) 間，然后再按這段時(shí)間在她愿意工作的總時(shí)間中所占的百分比來(lái)決定她的工資．現(xiàn)在請(qǐng)你幫約翰決定該雇傭哪些奶牛以保持牛棚的清潔，當(dāng)然，在能讓奶牛們滿意的 前提下，約翰希望使總花費(fèi)盡量小．

Input

* Line 1: Three space-separated integers: N, M, and E. * Lines 2..N+1: Line i+1 describes cow i's schedule with three space-separated integers: T1, T2, and S.

第1行：3個(gè)正整數(shù)N，M，E，用空格隔開(kāi)． 第2到N+1行：第i+l行給出了編號(hào)為i的奶牛的工作計(jì)劃，即3個(gè)用空格隔開(kāi)的正整數(shù)Ti，T2，S．

Output

* Line 1: a single integer that is either the minimum total salary to get the barn cleaned or else -1 if it is impossible to clean the barn.

輸出一個(gè)整數(shù)，表示約翰需要為牛棚清理工作支付的最少費(fèi)用．如果清理工作不可能完成， 那么輸出-1.

Sample Input

3 0 4 //三頭牛,要打掃從0到4號(hào)stall
0 2 3 //一號(hào)牛,從0號(hào)stall打掃到2號(hào),工資為3
3 4 2
0 0 1

INPUT DETAILS:

FJ has three cows, and the barn needs to be cleaned from second 0 to second
4. The first cow is willing to work during seconds 0, 1, and 2 for a total
salary of 3, etc.

Sample Output

5

HINT

約翰有3頭牛，牛棚在第0秒到第4秒之間需要打掃．第1頭牛想要在第0，1，2秒內(nèi)工作，為此她要求的報(bào)酬是3美元．其余的依此類推．????約翰雇傭前兩頭牛清掃牛棚，可以只花5美元就完成一整天的清掃．

Source

Silver

總結(jié)

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