POJ 1769 Minimizing maximizer （dp + 线段树）-CFANZ编程社区

题意：

找最少的区间使得依次连续覆盖所有n 个数。

思路：

令dp[i]表示覆盖到以i 为终点的区间的最少个数。

那么dp[i] 转移肯定来自 s[j]~t[j] 里面的，我们需要找一个k 使得 s[j] <= k <= t[j] 中 dp[k]最小值。

dp[i] = min(dp[i], dp[k]+1);

找最小值可以用线段树优化。

边界是dp[1] = 0

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int n, m;

int dp[50000 + 7];
int MIN[50000+7<<2];

const int inf = 0x3f3f3f3f;



void build(int l,int r,int o){
    if (l == r){
        MIN[o] = dp[l];
        return;
    }
    int m = l+r>>1;
    build(l,m,o<<1);
    build(m+1,r,o<<1|1);
    MIN[o] = min(MIN[o<<1], MIN[o<<1|1]);
}

int query(int L,int R,int l,int r,int o){
    if (L <= l && r <= R){
        return MIN[o];
    }

    int m = l+r>>1;
    int ans = inf;
    if (m >= L) ans = min(ans,query(L,R,l,m,o<<1));
    if (m < R) ans = min(ans,query(L,R,m+1,r,o<<1|1));
    return ans;
}


void update(int p,int v,int l,int r,int o){
    if (l == r){
        MIN[o] = v;
        return;
    }
    int m = l+r>>1;
    if (p <= m)update(p,v,l,m,o<<1);
    else update(p,v,m+1,r,o<<1|1);
    MIN[o] = min(MIN[o<<1],MIN[o<<1|1]);
}

int main(){
    scanf("%d %d",&n, &m);
    memset(dp,inf,sizeof dp);
    dp[1] = 0;
    build(1,n,1);
    for (int i = 0; i < m; ++i){
        int u,v;
        scanf("%d %d",&u, &v);
        int qq = query(u,v,1,n,1);
        if (qq+1 < dp[v]){
            dp[v] = qq+1;
            update(v,dp[v],1,n,1);
        }
    }
    printf("%d\n",dp[n]);
    return 0;
}

/**
40 6
20 30
1 10
10 20
20 30
15 25
30 40

ans = 4;

**/

Minimizing maximizer Description The company Chris Ltd. is preparing a new sorting hardware called Maximizer. Maximizer has n inputs numbered from 1 to n. Each input represents one integer. Maximizer has one output which represents the maximum value present on Maximizer's inputs. Maximizer is implemented as a pipeline of sorters Sorter(i1, j1), ... , Sorter(ik, jk). Each sorter has n inputs and n outputs. Sorter(i, j) sorts values on inputs i, i+1,... , j in non-decreasing order and lets the other inputs pass through unchanged. The n-th output of the last sorter is the output of the Maximizer. An intern (a former ACM contestant) observed that some sorters could be excluded from the pipeline and Maximizer would still produce the correct result. What is the length of the shortest subsequence of the given sequence of sorters in the pipeline still producing correct results for all possible combinations of input values? Task Write a program that: reads a description of a Maximizer, i.e. the initial sequence of sorters in the pipeline, computes the length of the shortest subsequence of the initial sequence of sorters still producing correct results for all possible input data, writes the result. Input The first line of the input contains two integers n and m (2 <= n <= 50000, 1 <= m <= 500000) separated by a single space. Integer n is the number of inputs and integer m is the number of sorters in the pipeline. The initial sequence of sorters is described in the next m lines. The k-th of these lines contains the parameters of the k-th sorter: two integers ik and jk (1 <= ik < jk <= n) separated by a single space. Output The output consists of only one line containing an integer equal to the length of the shortest subsequence of the initial sequence of sorters still producing correct results for all possible data. Sample Input 40 620 301 1010 2020 3015 2530 40 Sample Output 4 Hint Huge input data, scanf is recommended. Source Central Europe 2003