Codeforces Round #210 (Div. 2) C. Levko and Array Recovery

题目链接

线段树的逆过程，想了老一会，然后发现应该是包含区间对存在有影响，就不知怎么做了。。。然后尚大神，说，So easy，你要倒着来，然后再正着来，判断是不是合法就行了。然后我乱写了写，就过了。数据难道又水了。。。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
int p[5001],o[5001],ans[5001];
int t[5001],l[5001],r[5001],d[5001];
int main()
{
    int n,m,i,j,flag,maxz;
    scanf("%d%d",&n,&m);
    for(i = 1;i <= n;i ++)
    p[i] = -10000000;
    for(i = 1;i <= m;i ++)
    {
        scanf("%d%d%d%d",&t[i],&l[i],&r[i],&d[i]);
    }
    flag = 0;
    for(i = m;i >= 1;i --)
    {
        if(t[i] == 1)
        {
            for(j = l[i];j <= r[i];j ++)
            p[j] -= d[i];
        }
        else
        {
            for(j = l[i];j <= r[i];j ++)
            {
                if(o[j]&&p[j] < d[i]) continue;
                p[j] = d[i];
                o[j] = 1;
            }
        }
    }
    for(i = 1;i <= n;i ++)
    ans[i] = p[i];
    for(i = 1;i <= m;i ++)
    {
        if(t[i] == 1)
        {
            for(j = l[i];j <= r[i];j ++)
            p[j] += d[i];
        }
        else
        {
            maxz = -1000000000;
            for(j = l[i];j <= r[i];j ++)
            {
                maxz = max(maxz,p[j]);
            }
            if(maxz != d[i]) flag = 1;
        }
    }
    if(flag)
    printf("NO
");
    else
    {
        printf("YES
");
        for(i = 1;i <= n;i ++)
        {
            if(i == 1)
            printf("%d",ans[i]);
            else
            printf(" %d",ans[i]);
        }
        printf("
");
    }
    return 0;
}