HDU 1160 FatMouse's Speed LIS DP

http://acm.hdu.edu.cn/showproblem.php?pid=1160

同样是先按它的体重由小到大排，相同就按speed排就行。

这样做的好处是，能用O(n^2)枚举，因为前面的肯定不能和后面的搭配了。

然后就是LIS的题了，

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn =  + ;
struct node {
    int weight, speed;
    int id;
    node(int a, int b, int c) : weight(a), speed(b), id(c) {}
    node() {}
    bool operator < (const struct node & rhs) const {
        if (weight != rhs.weight) return weight < rhs.weight;
        else return speed > rhs.speed;
    }
}a[maxn];
struct DP {
    int val;
    struct node cur;
    int pre;
}dp[maxn];
bool isok(struct node a, struct node b) {
    if (a.weight < b.weight && a.speed > b.speed) return true;
    else return false;
}
bool isbetter(struct node a, struct node b) {
    if (a.weight != b.weight) return a.weight < b.weight;
    else if (a.speed != b.speed) return a.speed > b.speed;
    return false;
}
void show(int id) {
    if (id == -inf) return;
    show(dp[id].pre);
    printf("%d\n", a[id].id);
}
void work() {
    int total = ;
    int c, d;
    while (scanf("%d%d", &c, &d) != EOF) {
        ++total;
        a[total] = node(c, d, total);
    }
    sort(a + , a +  + total);
    for (int i = ; i <= total; ++i) {
        dp[i].val = ;
        dp[i].pre = -inf;
        dp[i].cur = a[i];
        for (int j = ; j < i; ++j) {
            if (isok(dp[j].cur, dp[i].cur)) { // i能接在j后面
                if (dp[i].val < dp[j].val + ) {
                    dp[i].val = dp[j].val + ;
                    dp[i].pre = j;
                } else if (dp[i].val == dp[j].val + ) {
                    if (isbetter(a[j], a[dp[i].pre])) { //j比它前一个好
                        dp[i].pre = j;
                    }
                }
            }
        }
    }
    int ans = -inf;
    int id;
    for (int i = ; i <= total; ++i) {
        if (ans < dp[i].val) {
            ans = dp[i].val;
            id = i;
        }
    }
    printf("%d\n", ans);
    show(id);
}

int main() {
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    work();
    return ;
}