1445: Pku3245 Sequence Partitioning

1445: Pku3245 Sequence Partitioning

Time Limit: 8 Sec  Memory Limit: 64 MB
Submit: 102  Solved: 59
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Description

Input

Output

Sample Input

4 6
4 3
3 5
2 5
2 4

Sample Output

9

HINT

An available assignment is the first two pairs are assigned into the first part and the last two pairs are assigned into the second part. Then B1 > A3, B1 > A4, B2 > A3, B2 > A4, max{A1, A2}+max{A3, A4} ≤ 6, and minimum max{B1+B2, B3+B4}=9.

Source

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对于三类限制条件，逐一分析对于第一条，转变成，如果有i < j && bi <= aj，那么从i到j这一段的所有物品必须分在同一组这样将必须合并的合并好，剩下的物品无论怎样划分都不会和第一个限制冲突题目要求第三类限制尽可能小，那么二分一个答案定义状态fi:前i个物品在当前二分答案下能够造出的合法方案的∑Mk的最小值那么有，fi = max{fj + max{Aj+1 ~ Ai}}由于每个位置i能够转移过来的j一定小于它并且，对于任意i < j，总有fi <= fj那么在从左往右扫描的时候，维护一个A单调递减的栈对于相邻两个元素x,y，从x ~ y-1这一段的转移增加的M的值都是Ay因为x是最左边的那个，所以用x转移肯定是最优的这样相邻的两个元素就用一个堆维护转移贡献对于最左端点，人工特判一下就行了

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#include<vector>#include<queue>#include<set>#include<map>#include<stack>#include<bitset>#include<ext/pb_ds/priority_queue.hpp>using namespace std; const int maxn = 1E5 + 10;typedef long long LL;const LL INF = 1E15; struct data{    int Num; LL F,G; data(){}    data(int Num,LL F,LL G): Num(Num),F(F),G(G){}    bool operator < (const data &B) const {return F + G > B.F + B.G;}}; int n,cnt,head,tail,q[maxn],A[maxn],B[maxn],R[maxn],Max[maxn];LL m,sum[maxn],f[maxn],g[maxn];bool inq[maxn]; priority_queue <data> Q; void Add(int l,int r){    ++cnt; Max[cnt] = 0;    for (int i = l; i <= r; i++)        sum[cnt] += 1LL * B[i],Max[cnt] = max(Max[cnt],A[i]);} LL Get_top(){    while (!Q.empty())    {        data k = Q.top();        if (!inq[k.Num] || g[k.Num] != k.G) {Q.pop(); continue;}        else return k.F + k.G;    }    return INF + 233LL;} bool Judge(LL now){    int tmp = q[head = tail = 1] = 0;    while (!Q.empty()) Q.pop();    Q.push(data(0,0,0)); inq[0] = 1;    for (int i = 1; i <= cnt; i++)    {        f[i] = INF; g[i] = 0;        while (sum[i] - sum[tmp] > now) ++tmp;        while (head <= tail && q[head] < tmp)            inq[q[head]] = 0,++head;        while (head <= tail && Max[q[tail]] <= Max[i])            inq[q[tail]] = 0,--tail;        if (head <= tail)        {            g[q[tail]] = Max[i]; inq[q[tail]] = 1;            Q.push(data(q[tail],f[q[tail]],g[q[tail]]));        }        f[i] = Get_top(); q[++tail] = i;        f[i] = min(f[i],f[tmp] + 1LL * Max[q[head]]);    }    return f[cnt] <= m;} int getint(){    char ch = getchar(); int ret = 0;    while (ch < '0' || '9' < ch) ch = getchar();    while ('0' <= ch && ch <= '9')        ret = ret * 10 + ch - '0',ch = getchar();    return ret;} int main(){    #ifdef DMC        freopen("DMC.txt","r",stdin);    #endif         cin >> n >> m;    for (int i = 1; i <= n; i++)        Max[i] = A[i] = getint(),B[i] = getint();    for (int i = n - 1; i; i--) Max[i] = max(Max[i],Max[i + 1]);    for (int i = 1; i <= n; i++)    {        int l = i,r = n;        while (r - l > 1)        {            int mid = l + r >> 1;            if (Max[mid] >= B[i]) l = mid;            else r = mid;        }        R[i] = Max[r] >= B[i] ? r : l;    }    int tl = 1,tr = R[1];    for (int i = 2; i <= n; i++)        if (tr < i) {Add(tl,tr); tl = i; tr = R[i];}        else tr = max(tr,R[i]); Add(tl,tr);         for (int i = 1; i <= cnt; i++) sum[i] += sum[i - 1];    LL l,r; l = 0; r = sum[cnt]; Max[0] = ~0U>>1;    while (r - l > 1LL)    {        LL mid = l + r >> 1LL;        if (Judge(mid)) r = mid;        else l = mid;    }    cout << (Judge(l) ? l : r) << endl;    return 0;}