題干：

Arkady plays Gardenscapes a lot. Arkady wants to build two new fountains. There are?n?available fountains, for each fountain its beauty and cost are known. There are two types of money in the game: coins and diamonds, so each fountain cost can be either in coins or diamonds. No money changes between the types are allowed.

Help Arkady to find two fountains with maximum total beauty so that he can buy both at the same time.

Input

The first line contains three integers?n,?c?and?d?(2?≤?n?≤?100?000,?0?≤?c,?d?≤?100?000)?— the number of fountains, the number of coins and diamonds Arkady has.

The next?n?lines describe fountains. Each of these lines contain two integers?bi?and?pi?(1?≤?bi,?pi?≤?100?000)?— the beauty and the cost of the?i-th fountain, and then a letter "C" or "D", describing in which type of money is the cost of fountain?i: in coins or in diamonds, respectively.

Output

Print the maximum total beauty of exactly two fountains Arkady can build. If he can't build two fountains, print?0.

Examples

Input

3 7 6 10 8 C 4 3 C 5 6 D

Output

9

Input

2 4 5 2 5 C 2 1 D

Output

0

Input

3 10 10 5 5 C 5 5 C 10 11 D

Output

10

Note

In the first example Arkady should build the second fountain with beauty?4, which costs?3?coins. The first fountain he can't build because he don't have enough coins. Also Arkady should build the third fountain with beauty?5?which costs?6?diamonds. Thus the total beauty of built fountains is?9.

In the second example there are two fountains, but Arkady can't build both of them, because he needs?5?coins for the first fountain, and Arkady has only?4?coins.

題目大意：

?

解題報(bào)告：

? ?比賽的時(shí)候讓這個(gè)線(xiàn)段樹(shù)搞垮我了，，很多特殊情況要考慮，并且，最后卡住我的不是線(xiàn)段樹(shù)那一塊，而是之前的C和D兩種貨幣各選一個(gè)的那種情況，我是直接，找到就break，但是其實(shí)應(yīng)該是維護(hù)一個(gè)最大值。。。因?yàn)槟阒苯觔reak的話(huà)只能保證cost是可符合的最大的，而保證不了val的最大性！我們需要的是val！（%%%WJH大佬）

AC代碼1：

#include<bits/stdc++.h> using namespace std; const int MAX = 100000 +5; int n,c,d,totc,totd; int cc[MAX]; int dd[MAX]; struct Node{int val,cost;Node(){}Node(int val,int cost):val(val),cost(cost){} } C[MAX],D[MAX]; struct TREE {int l,r;int val; } tree[MAX*4]; bool cmp(const Node &a,const Node &b) {return a.cost < b.cost; } void pushup(int cur) {tree[cur].val = max(tree[cur*2].val,tree[cur*2+1].val); } void buildc(int l,int r,int cur) {tree[cur].l=l;tree[cur].r=r;if(l == r) {tree[cur].val = C[l].val;return ;}int m = (l +r)/2;buildc(l,m,cur*2);buildc(m+1,r,cur*2+1);pushup(cur); } void buildd(int l,int r,int cur) {tree[cur].l=l;tree[cur].r=r;if(l == r) {tree[cur].val = D[l].val;return ;}int m = (l +r)/2;buildd(l,m,cur*2);buildd(m+1,r,cur*2+1);pushup(cur); } int query(int pl,int pr,int cur) {if(pl <= tree[cur].l && pr >= tree[cur].r) return tree[cur].val;int res = 0;if(pl <= tree[cur*2].r) res = query(pl,pr,cur*2);if(pr >= tree[cur*2+1].l) res = max(res,query(pl,pr,cur*2+1));return res; } void update(int tar,int val,int cur) {if(tree[cur].l == tree[cur].r) {tree[cur].val = val;return ;}int m = (tree[cur].l + tree[cur].r)/2;if(tar <= m) update(tar,val,cur*2);else update(tar,val,cur*2+1);pushup(cur); } signed main() {char op[10];int val,cost,flag1=0,flag2=0;cin>>n>>c>>d;for(int i = 1; i<=n; i++) {scanf("%d%d%s",&val,&cost,op);if(op[0] == 'C') {C[++totc] = Node(val,cost);}else {D[++totd] = Node(val,cost);}}sort(C+1,C+totc+1,cmp); sort(D+1,D+totd+1,cmp);int maxx=0,tmp1 = 0,tmp2 = 0,pos;for(int i = totc; i>=1; i--) {if(C[i].cost <= c) {flag1=1;tmp1 = max(tmp1,C[i].val);}}for(int i = totd; i>=1; i--) {if(D[i].cost <= d) {flag2=1;tmp2 = max(tmp2,D[i].val);}}if(flag1==0 && flag2 == 0) {printf("0\n");return 0;} if(flag1 == 1 && flag2 == 1) {maxx = tmp1 + tmp2;}if(totc == 1 && totd == 1) {if(C[1].cost > c || D[1].cost > d) {printf("0\n");return 0;}}for(int i = 1; i<=totc; i++) cc[i] = C[i].cost;for(int i = 1; i<=totd; i++) dd[i] = D[i].cost; // printf("maxx = %d\n",maxx);//處理Cif(totc != 0 ) {buildc(1,totc,1);for(int i =1; i<=totc; i++) {if(c-cc[i] < 0) break; update(i,0,1);pos = upper_bound(cc+1,cc+totc + 1,c-cc[i]) - cc-1;if(pos == 0) continue;maxx = max(maxx,C[i].val + query(1,pos,1));update(i,C[i].val,1);}}for(int i = 1; i<=MAX; i++) update(i,0,1); // printf("%d %d\n",query(1,1,1),query(1,2,1));if(totd !=0) {buildd(1,totd,1);for(int i =1; i<=totd; i++) {if(d-dd[i] < 0) break;update(i,0,1);pos = upper_bound(dd+1,dd+totd + 1,d-dd[i])- dd -1;if(pos == 0) continue;// if(dd[pos] > d-dd[i]) pos--;maxx = max(maxx,D[i].val + query(1,pos,1));update(i,D[i].val,1);}}printf("%d\n",maxx);return 0 ; } //6 4 9 //6 6 D //1 4 D //6 7 C //7 6 D //5 7 D //2 5 D//6 68 40 //1 18 D //6 16 D //11 16 D //7 23 D //16 30 D //2 20 D//3 9 5 //10 7 C //8 6 C //5 3 C

AC代碼2：（o(n^2)）

#include<bits/stdc++.h> #define ll long long using namespace std; const int MAX = 1e5 + 5 ; int n,c,d,ans; struct Node {int cost,val;char op[5]; } node[MAX]; bool cmp(const Node & a,const Node & b) {if(a.val != b.val) return a.val > b.val;return a.cost < b.cost; } int main() {cin>>n>>c>>d;for(int i = 1; i<=n; i++) {scanf("%d%d%s",&node[i].val,&node[i].cost,node[i].op);}sort(node+1,node+n+1,cmp);int cc,dd,flag,tmp;for(int i = 1; i<=n; i++) {cc = c;dd = d;flag = 0 ;if(node[i].op[0] == 'C' && node[i].cost <= c) {flag=1;cc-=node[i].cost;tmp=node[i].val;}else if(node[i].op[0] == 'D' && node[i].cost <= d) {flag=1;dd-=node[i].cost;tmp = node[i].val;}if(flag == 0) continue;for(int j = i+1; j<=n; j++) {if(node[j].op[0] == 'C' && node[j].cost <= cc) {ans = max(ans,tmp+node[j].val);break;}else if(node[j].op[0] == 'D' && node[j].cost <= dd) {ans = max(ans,tmp+node[j].val);break;}}}printf("%d\n",ans);return 0 ; }

AC代碼3：（樹(shù)狀數(shù)組）

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm>using namespace std;const int maxn = 100000;int C[maxn+10],D[maxn+10];void add(int *tree,int k,int num) {while(k<=maxn){tree[k] = max(tree[k],num);k+=k&-k;} }int read(int *tree,int k) {int res=0;while(k){res = max(res,tree[k]);k-=k&-k;}return res; }int main(void) {int n,c,d,i,j;while(scanf("%d%d%d",&n,&c,&d)==3){memset(C,0,sizeof(C));memset(D,0,sizeof(D));int ans = 0;for(i=1;i<=n;i++){int b,p;char t[5];scanf("%d%d%s",&b,&p,t);int maxn;if(t[0] == 'C'){maxn = read(D,d);if(p > c)continue;maxn = max(maxn,read(C,c-p));add(C,p,b);}else{maxn = read(C,c);if(p > d)continue;maxn = max(maxn,read(D,d-p));add(D,p,b);}if(maxn)ans = max(ans,maxn + b);}cout << ans << endl;}return 0; }

AC代碼4：（tourist）

#include <bits/stdc++.h>using namespace std;const int N = 1234567;int pref[N], a[N], b[N]; char c[N];int solve(vector < pair <int, int> > z, int b) {sort(z.begin(), z.end());int cnt = z.size();if (cnt < 2) {return 0;}pref[0] = 0;for (int i = 0; i < cnt; i++) {pref[i + 1] = max(pref[i], z[i].second);}int i = 0;int res = 0;for (int j = cnt - 1; j >= 0; j--) {while (i < j && z[i].first + z[j].first <= b) {i++;}i = min(i, j);if (i > 0) {res = max(res, pref[i] + z[j].second);}}return res; }int main() {int n, C, D;scanf("%d %d %d", &n, &C, &D);for (int i = 0; i < n; i++) {scanf("%d %d", a + i, b + i);c[i] = getchar();while (c[i] != 'C' && c[i] != 'D') {c[i] = getchar();}}int ans = 0;{int x = 0, y = 0;for (int i = 0; i < n; i++) {if (c[i] == 'C' && b[i] <= C) {x = max(x, a[i]);}}for (int i = 0; i < n; i++) {if (c[i] == 'D' && b[i] <= D) {y = max(y, a[i]);}}if (x > 0 && y > 0) {ans = max(ans, x + y);}}{vector < pair <int, int> > z;for (int i = 0; i < n; i++) {if (c[i] == 'C') {z.push_back(make_pair(b[i], a[i]));}}ans = max(ans, solve(z, C));}{vector < pair <int, int> > z;for (int i = 0; i < n; i++) {if (c[i] == 'D') {z.push_back(make_pair(b[i], a[i]));}}ans = max(ans, solve(z, D));}printf("%d\n", ans);return 0; }

AC代碼5：（在線(xiàn)搞，以cost建樹(shù)（而不是排好序后用下標(biāo)建樹(shù)），維護(hù)最大值）

#include<cstdio> #include<algorithm> #include<cstring> #define ll long long using namespace std; const int maxn = 100010; int a[maxn],acost[maxn],b[maxn],bcost[maxn]; struct node {int Max;int l,r; }t[maxn*4]; void build(int root,int l,int r) {t[root].l=l;t[root].r=r;//t[root].Max=-1;if(l==r)return;int mid=(l+r)>>1;build(2*root,l,mid);build(2*root+1,mid+1,r); } void update(int root,int index,int val) {int l=t[root].l,r=t[root].r;if(l==r){t[root].Max=max(t[root].Max,val);return;//}int mid=(l+r)>>1;if(index<=mid)update(2*root,index,val);elseupdate(2*root+1,index,val);t[root].Max=max(t[root*2].Max,t[root*2+1].Max); } int query(int root,int ql,int qr) {int l=t[root].l;int r=t[root].r;if(l==ql&&r==qr){return t[root].Max;}int mid=(l+r)>>1;if(qr<=mid)return query(2*root,ql,qr);else if(ql>mid)return query(2*root+1,ql,qr);elsereturn max(query(2*root,ql,mid),query(2*root+1,mid+1,qr));//2*root+1 } int main() {int N,C,D;while(~scanf("%d%d%d",&N,&C,&D)){int acnt=0,bcnt=0;char ch;int c,w;int Maxc=-1;for(int i=0;i<N;i++){scanf("%d%d %c",&c,&w,&ch);if(ch=='C'){a[acnt]=c;acost[acnt++]=w;}else if(ch=='D'){b[bcnt]=c;bcost[bcnt++]=w;}Maxc=max(w,Maxc);}int ans=-1;int res1=-1,res2=-1;for(int i=0;i<acnt;i++){if(C>=acost[i])res1=max(res1,a[i]);}for(int i=0;i<bcnt;i++){if(D>=bcost[i])res2=max(res2,b[i]);}if(res1!=-1&&res2!=-1)ans=res1+res2;Maxc=max(Maxc,max(C,D));build(1,1,Maxc+10);for(int i=0;i<acnt;i++){int res=-1;if(C>acost[i])res=query(1,1,C-acost[i]);update(1,acost[i],a[i]);if(res!=-1)ans=max(ans,a[i]+res);}build(1,1,Maxc);for(int i=0;i<bcnt;i++){int res=-1;if(D>bcost[i])res=query(1,1,D-bcost[i]);update(1,bcost[i],b[i]);if(res!=-1)ans=max(ans,b[i]+res);}if(ans!=-1)printf("%d\n",ans);elseprintf("0\n");} }

AC代碼6：（維護(hù)最大價(jià)值和次大價(jià)值 + 前綴和）（感覺(jué)這個(gè)題解有問(wèn)題？）

#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define rep(i,a,b) for (int i=a; i<=b; ++i) #define per(i,b,a) for (int i=b; i>=a; --i) #define mes(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f #define MP make_pair #define PB push_back #define fi first #define se second #define PII pair<ll , ll > typedef long long ll; const int N = 100005;int n; ll C, D; PII c[N], d[N]; ll ans1[N][2], ans2[N][2]; int cnt1=0, cnt2=0; //init函數(shù)中均可加等號(hào) void Init() {sort(c+1, c+1+cnt1);sort(d+1, d+1+cnt2);ll mx1=0, mx2=0;rep(i,1,cnt1){if(mx1 < c[i].se){mx2=mx1, mx1=c[i].se;ans1[c[i].fi][0]=mx1;ans1[c[i].fi][1]=mx2;}else if(mx2 < c[i].se){mx2=c[i].se;ans1[c[i].fi][0]=mx1;//可注釋ans1[c[i].fi][1]=mx2;}}mx1=0, mx2=0;rep(i,1,cnt2){if(mx1 < d[i].se){mx2=mx1, mx1=d[i].se;ans2[d[i].fi][0]=mx1;ans2[d[i].fi][1]=mx2;}else if(mx2 < d[i].se){mx2=d[i].se;ans2[d[i].fi][0]=mx1;//可注釋ans2[d[i].fi][1]=mx2;}}rep(i,1,max(C,D)) rep(j,0,1){ans1[i][j]=max(ans1[i][j], ans1[i-1][j]);ans2[i][j]=max(ans2[i][j], ans2[i-1][j]);} } int main() {scanf("%d%lld%lld", &n, &C, &D);ll bi, pi; char ch;rep(i,1,n){scanf("%lld%lld%*c%c", &bi, &pi, &ch);if(ch=='C') c[++cnt1]=MP(pi, bi);else d[++cnt2]=MP(pi, bi);}Init();ll sum1=0;if(ans1[C][0] && ans2[D][0])sum1 = max(sum1, ans1[C][0]+ans2[D][0]);ll sum2=0, sum3=0;rep(i,1,cnt1){ll ret=C-c[i].fi;if(ret>0){if(ans1[ret][0]==c[i].se){if(ans1[ret][1])sum2 = max(sum2, c[i].se+ans1[ret][1]);}else{if(ans1[ret][0])sum2 = max(sum2, c[i].se+ans1[ret][0]);}}}rep(i,1,cnt2){ll ret=D-d[i].fi;if(ret>0){if(ans2[ret][0]==d[i].se){if(ans2[ret][1])sum3 = max(sum3, d[i].se+ans2[ret][1]);}else{if(ans2[ret][0])sum3 = max(sum3, d[i].se+ans2[ret][0]);}}}printf("%lld\n", max(sum1, max(sum2, sum3)));return 0; }

AC代碼7：（還是自己寫(xiě)的那個(gè)線(xiàn)段樹(shù)，只是將upperbound部分進(jìn)行了優(yōu)化，快了將近50ms）

#include<bits/stdc++.h> using namespace std; const int MAX = 100000 +5; int n,c,d,totc,totd; int cc[MAX]; int dd[MAX]; struct Node{int val,cost;Node(){}Node(int val,int cost):val(val),cost(cost){} } C[MAX],D[MAX]; struct TREE {int l,r;int val; } tree[MAX*4]; bool cmp(const Node &a,const Node &b) {if(a.cost != b.cost)return a.cost < b.cost;return a.val > b.val; } void pushup(int cur) {tree[cur].val = max(tree[cur*2].val,tree[cur*2+1].val); } void buildc(int l,int r,int cur) {tree[cur].l=l;tree[cur].r=r;if(l == r) {tree[cur].val = C[l].val;return ;}int m = (l +r)/2;buildc(l,m,cur*2);buildc(m+1,r,cur*2+1);pushup(cur); } void buildd(int l,int r,int cur) {tree[cur].l=l;tree[cur].r=r;if(l == r) {tree[cur].val = D[l].val;return ;}int m = (l +r)/2;buildd(l,m,cur*2);buildd(m+1,r,cur*2+1);pushup(cur); } int query(int pl,int pr,int cur) {if(pl <= tree[cur].l && pr >= tree[cur].r) return tree[cur].val;int res = 0;if(pl <= tree[cur*2].r) res = query(pl,pr,cur*2);if(pr >= tree[cur*2+1].l) res = max(res,query(pl,pr,cur*2+1));return res; } void update(int tar,int val,int cur) {if(tree[cur].l == tree[cur].r) {tree[cur].val = val;return ;}int m = (tree[cur].l + tree[cur].r)/2;if(tar <= m) update(tar,val,cur*2);else update(tar,val,cur*2+1);pushup(cur); } int main() {char op[10];int val,cost,flag1=0,flag2=0;cin>>n>>c>>d;for(int i = 1; i<=n; i++) {scanf("%d%d%s",&val,&cost,op);if(op[0] == 'C') {C[++totc] = Node(val,cost);}else {D[++totd] = Node(val,cost);}}sort(C+1,C+totc+1,cmp); sort(D+1,D+totd+1,cmp);int maxx=0,tmp1 = 0,tmp2 = 0,pos;for(int i = totc; i>=1; i--) {if(C[i].cost <= c) {flag1=1;tmp1 = max(tmp1,C[i].val);}}for(int i = totd; i>=1; i--) {if(D[i].cost <= d) {flag2=1;tmp2 = max(tmp2,D[i].val);}}if(flag1==0 && flag2 == 0) {printf("0\n");return 0;} if(flag1 == 1 && flag2 == 1) {maxx = tmp1 + tmp2;}if(totc == 1 && totd == 1) {if(C[1].cost > c || D[1].cost > d) {printf("0\n");return 0;}}for(int i = 1; i<=totc; i++) cc[i] = C[i].cost;for(int i = 1; i<=totd; i++) dd[i] = D[i].cost; // printf("maxx = %d\n",maxx);//處理Cif(totc != 0 ) {buildc(1,totc,1);for(int i =1; i<totc; i++) {if(c-cc[i] < 0) break; // update(i,0,1);pos = upper_bound(cc+i+1,cc+totc + 1,c-cc[i]) - (cc/*+i*/)-1;if(pos == i) continue;maxx = max(maxx,C[i].val + query(i+1,pos,1)); // update(i,C[i].val,1);}} // printf("%d %d\n",query(1,1,1),query(1,2,1));if(totd !=0) {buildd(1,totd,1);for(int i =1; i<totd; i++) {if(d-dd[i] < 0) break; // update(i,0,1);pos = upper_bound(dd+i+1,dd+totd + 1,d-dd[i])- (dd/*+i*/) -1;if(pos == i) continue;// if(dd[pos] > d-dd[i]) pos--;maxx = max(maxx,D[i].val + query(i+1,pos,1)); // update(i,D[i].val,1);}}printf("%d\n",maxx);return 0 ; } //6 4 9 //6 6 D //1 4 D //6 7 C //7 6 D //5 7 D //2 5 D//6 68 40 //1 18 D //6 16 D //11 16 D //7 23 D //16 30 D //2 20 D//3 9 5 //10 7 C //8 6 C //5 3 C

總結(jié)：

? ? 對(duì)于這一道經(jīng)典的題目，我想，可以總結(jié)的東西還是很多的。

? ?對(duì)于任何數(shù)據(jù)結(jié)構(gòu)（包括線(xiàn)段樹(shù)等，包括棧隊(duì)列等），首先需要看的就是能否進(jìn)入，或者說(shuō)，這項(xiàng)功能能否執(zhí)行? 的問(wèn)題。比如棧，進(jìn)行pop和top操作的時(shí)候要想，是否需要先判斷是否棧為空呢？因?yàn)闉榭樟司蜎](méi)法進(jìn)行操作了呀。同理，對(duì)于線(xiàn)段樹(shù)，我們也要想，我們能否建樹(shù)呢？建樹(shù)后，能否進(jìn)行查詢(xún)呢？即，這個(gè)傳參是合法的嗎？（這兩條思考，分別對(duì)應(yīng)了AC代碼1中的，if(totc!=0),,,,if(c-cc[i] < 0) break; 這兩句）

? ?其次是這一塊的處理，也經(jīng)常容易忽略：（一個(gè)比較可行的解決辦法就是自己造數(shù)據(jù)的時(shí)候這些邊界情況都考慮一下）（好像說(shuō)了一堆廢話(huà)）

if(flag1==0 && flag2 == 0) {printf("0\n");return 0;} if(flag1 == 1 && flag2 == 1) {maxx = tmp1 + tmp2;}if(totc == 1 && totd == 1) {if(C[1].cost > c || D[1].cost > d) {printf("0\n");return 0;}}

將ac代碼1修改成AC代碼7：

? ? ?改1：for(int i =1; i<totc; i++) {

? ? ?改2：if(pos == i) continue;

? ? ?改3：maxx = max(maxx,C[i].val + query(i+1,pos,1));//這里query中是從i+1開(kāi)始的

? ? ?改4：每個(gè)for中都可以注釋掉兩個(gè)update函數(shù)。（因?yàn)榫蜎](méi)用了）

?

還有一些可總結(jié)的：

? ? lowerbound? upperbound的時(shí)候，如果不是對(duì)整個(gè)數(shù)列進(jìn)行，比如是對(duì)[ l , r ]進(jìn)行，那么upper的可能值就是[?l , r+1 ]這個(gè)區(qū)間內(nèi)的值。（即upper出來(lái)的pos，肯定大于等于l，小于等于r+1），同理 lowerbound出來(lái)的值略有不同，區(qū)間是[ l , r ]（我感覺(jué)是這樣？）

? ?還有就是，對(duì)[ l , r ]進(jìn)行的時(shí)候，但是還是要 -a? 而不是-(a+i-1)? 也就是int pos = lower_bound(a+i,a+i+1,val) - a;這樣就可以，因?yàn)槟阕詈笮枰倪€是在原數(shù)組中的下標(biāo)呀。

? 總之做題還是要在仔細(xì)一些吧。

另一個(gè)代碼:https://blog.csdn.net/SSimpLe_Y/article/details/71702915

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