dp

codeforces 4D

题意：给定n和信封和一个w 和 h，要求以下的信封选取最多的，然后 wi 必须比 w 大并递增，hi 比 h 大也递增，问最多的信封数和哪些，按照顺序输出

题解：按照w递增排序后，求h的最长上升子序列并输出路径

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
#include<cstdlib>
#include<queue>
#include<set>
#include<string.h>
#include<vector>
#include<deque>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define eps 1e-4
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define ll long long
#define LL long long
const int MAXN = 1e5 + 5;
const int mod = 998244353;

struct node{
    int w,h;
    int id;
}a[MAXN];
bool cmp(node a,node b) {
    if(a.w != b.w) return a.w < b.w;
    return a.h < b.h;
}
int dp[MAXN];
int path[MAXN];

void print(int n) {
    if(path[n] == -1) {
        printf("%d ",a[n].id);
        return ;
    } else {
        print(path[n]);
        printf("%d ",a[n].id);
    }
}
int main() {
    int n, w, h;
    cin >> n >> w >> h;
    int pos = 0;
    for(int i = 1; i <= n; i++) {
        int w1,h1;
        cin >> w1 >> h1;
        if(w1 > w && h1 > h) {
            a[pos].w = w1;
            a[pos].h = h1;
            a[pos].id = i;
            pos++;
        }
    }
    if(pos == 0)
    {
        cout << 0 << endl;
        return 0;
    }
    sort(a, a + pos, cmp);

    for(int i = 0; i < pos; i++) {
        dp[i] = 1;
        path[i] = i;
    }
    memset(path, -1, sizeof path);
    int maxlen = 1,biao = 0;
    for(int i = 0; i < pos; i++) {
        for(int j = 0; j < i; j++) {
            if(a[j].h < a[i].h && a[j].w < a[i].w && dp[i] < (dp[j] + 1)) {
                dp[i] = dp[j] + 1;
                if(dp[i] > maxlen) maxlen = dp[i];
                path[i] = j;
            }
        }
    }
    int tmp,cmp = 0;
    for(int i = 0; i  < n; i++) {
        if(dp[i] > cmp) {
            cmp = dp[i];
            tmp = i;
        }
    }
    cout << maxlen << endl;
//    debug(tmp);
    print(tmp);
}

 

来源：https://www.cnblogs.com/smallhester/p/11394118.html