poj 2373 Dividing the Path

很久之前做的了，已经想不起思路了，容我日后补充。

#include <stdio.h>
#include <deque>
#include <algorithm>
#define inf 1000000000
using namespace std;
int n, L, a, b, cnt = 1;
int f[1000010];
bool d[1000010];
struct node
{
    int s, e;
}p[1001], P[1001];
bool cmp(const node &a, const node &b)
{
    if(a.s != b.s) return a.s < b.s;
    else return a.e > b.e;
}
int main()
{
    int beg,end,i,j,t;
    scanf("%d%d%d%d", &n, &L, &a, &b);
    for(i = 1; i <= n; ++i)
        scanf("%d%d", &p[i].s, &p[i].e);
    sort(p + 1, p + n + 1, cmp);
    P[1] = p[1];
    for(i = 2; i <= n; ++i)
        if(p[i].e > p[i - 1].e)
            P[++cnt] = p[i];
    beg = end = 0;
    for(i = 1; i <= cnt; ++i)
    {
        if(P[i].s >= end)
        {
            for(j = beg + 1; j <= end - 1; ++j)
                d[j] = 1;
            beg = P[i].s;
            end = P[i].e;
        }
        else end = P[i].e;
    }
    for(j = beg + 1; j <= end - 1; ++j)
        d[j] = 1;
    f[0] = 0;
    deque<int>Q;
    for(i = 1; i <= L; ++i)
    {
        f[i] = inf;
        while(!Q.empty() && Q.front() < i - 2 * b)
            Q.pop_front();
        if(!d[i] && i % 2 == 0)
        {
            if(!Q.empty())
                f[i] = min(f[i], f[Q.front()] + 1);
        }
        t = i - 2 * a + 1;
        if(t >= 0 && !d[t] && t % 2 == 0)
        {
            while(!Q.empty() && f[t] < f[Q.back()])
                Q.pop_back();       
            Q.push_back(t);
        }
    }
    if(f[L] == inf) puts("-1");
    else printf("%d\n", f[L]);
    return 0;
}