POJ 1661 Help Jimmy （一个神trick）

小人从上往下掉的时候，比如区间[-50, 100], [0, 100]它可以从上一个100跳到下一个100.。。。。无语的神逻辑。。。从早晨wa到现在。T_T

按高度从大到小排序

f[i][0]表示到i层左端点的最小路径，f[i][1]表示到i层右端点的最小路径。

落下时判断第j层到第i层有没有其他的层。。。这个写了一个暴力，居然没有TLE。。。看来题目数据真的不怎么强。

View Code

#include <iostream>

#include <cstdio>

#include <cmath>

#include <vector>

#include <cstring>

#include <algorithm>

#include <string>

#include <set>

#include <ctime>

#include <queue>

#include <map>

#include <sstream>



#define CL(arr, val)    memset(arr, (val), sizeof(arr))

#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))

#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))

#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))

#define L(x)    (x) << 1

#define R(x)    (x) << 1 | 1

#define MID(l, r)   ((l) + (r)) >> 1

#define Min(x, y)   (x) < (y) ? (x) : (y)

#define Max(x, y)   (x) < (y) ? (y) : (x)

#define E(x)    (1 << (x))

#define iabs(x)  ((x) > 0 ? (x) : -(x))



typedef long long LL;

const double eps = 1e-8;

const int inf = ~0u>>2;



using namespace std;



const int N = 1024;



struct node {

    int x1, x2, h;

    bool operator < (const node tmp) const {

        return h > tmp.h;

    }

} l[N];



int f[N][2];



int main() {

    //freopen("data.in", "r", stdin);



    int t, n, x, y, mx;

    int i, j, k, tmp;

    bool flag;



    scanf("%d", &t);

    while(t--) {

        scanf("%d%d%d%d", &n, &x, &y, &mx);

        l[0].x1 = x; l[0].x2 = x; l[0].h = y;



        for(i = 1; i <= n; ++i) {

            scanf("%d%d%d", &l[i].x1, &l[i].x2, &l[i].h);

            f[i][1] = f[i][0] = inf;

        }

        sort(l, l + n + 1);

        f[0][0] = f[0][1] = 0;



        /*for(i = 0; i <= n; ++i) {

            printf("%d %d %d\n", l[i].x1, l[i].x2, l[i].h);

        }*/



        for(i = 1; i <= n; ++i) {

            for(j = 0; j < i; ++j) {

                if(l[i].x1 <= l[j].x1 && l[i].x2 >= l[j].x1 && l[j].h - l[i].h <= mx) {



                    //left

                    flag = true;

                    for(k = j + 1; k < i && flag; ++k)  {

                        if(l[k].x1 <= l[j].x1 && l[k].x2 >= l[j].x1)  {flag = false; break;}

                    }

                    if(flag && f[j][0] != inf) {

                        tmp = f[j][0] + l[j].h - l[i].h + iabs(l[j].x1 - l[i].x1);

                        f[i][0] = min(f[i][0], tmp);



                        tmp = f[j][0] + l[j].h - l[i].h + iabs(l[j].x1 - l[i].x2);

                        f[i][1] = min(f[i][1], tmp);

                    }

                }



                //right

                if(l[i].x1 <= l[j].x2 && l[i].x2 >= l[j].x2 && l[j].h - l[i].h <= mx) {

                    flag = true;

                    for(k = j + 1; k < i && flag; ++k) {

                        if(l[k].x1 <= l[j].x2 && l[k].x2 >= l[j].x2)  {flag = false; break;}

                    }

                    if(flag && f[j][1] != inf) {

                        tmp = f[j][1] + l[j].h - l[i].h + iabs(l[j].x2 - l[i].x1);

                        f[i][0] = min(f[i][0], tmp);



                        tmp = f[j][1] + l[j].h - l[i].h + iabs(l[j].x2 - l[i].x2);

                        f[i][1] = min(f[i][1], tmp);

                    }

                }

            }

        }



        int ans = inf;

        for(i = 0; i <= n; ++i) {

            if(l[i].h <= mx) {

                //left

                flag = true;

                for(k = i + 1; k <= n; ++k) {

                    if(l[k].x1 <= l[i].x1 && l[k].x2 >= l[i].x1)  {flag = false; break;}

                }

                if(flag) {

                    tmp = f[i][0] + l[i].h;

                    ans = min(tmp, ans);

                }



                //right

                flag = true;

                for(k = i + 1; k <= n; ++k) {

                    if(l[k].x1 < l[i].x2 && l[k].x2 > l[i].x2)  {flag = false; break;}

                }

                if(flag) {

                    ans = min(ans, f[i][1] + l[i].h);

                }

            }

        }

        printf("%d\n", ans);

    }

    return 0;

}