• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>

            hdu3440(差分約束)

            題目網(wǎng)址: http://acm.hdu.edu.cn/showproblem.php?pid=3440 

            House Man

            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
            Total Submission(s): 446    Accepted Submission(s): 157


            Problem Description
            In Fuzhou, there is a crazy super man. He can’t fly, but he could jump from housetop to housetop. Today he plans to use N houses to hone his house hopping skills. He will start at the shortest house and make N-1 jumps, with each jump taking him to a taller house than the one he is jumping from. When finished, he will have been on every house exactly once, traversing them in increasing order of height, and ending up on the tallest house.
            The man can travel for at most a certain horizontal distance D in a single jump. To make this as much fun as possible, the crazy man want to maximize the distance between the positions of the shortest house and the tallest house.
            The crazy super man have an ability—move houses. So he is going to move the houses subject to the following constraints:
            1. All houses are to be moved along a one-dimensional path.
            2. Houses must be moved at integer locations along the path, with no two houses at the same location.
            3. Houses must be arranged so their moved ordering from left to right is the same as their ordering in the input. They must NOT be sorted by height, or reordered in any way. They must be kept in their stated order.
            4. The super man can only jump so far, so every house must be moved close enough to the next taller house. Specifically, they must be no further than D apart on the ground (the difference in their heights doesn't matter).
            Given N houses, in a specified order, each with a distinct integer height, help the super man figure out the maximum possible distance they can put between the shortest house and the tallest house, and be able to use the houses for training.
             

            Input
            In the first line there is an integer T, indicates the number of test cases.(T<=500)
            Each test case begins with a line containing two integers N (1 ≤ N ≤ 1000) and D (1 ≤ D ≤1000000). The next line contains N integer, giving the heights of the N houses, in the order that they should be moved. Within a test case, all heights will be unique.
             

            Output
            For each test case , output “Case %d: “first where d is the case number counted from one, then output a single integer representing the maximum distance between the shortest and tallest house, subject to the constraints above, or -1 if it is impossible to lay out the houses. Do not print any blank lines between answers.
             

            Sample Input
            3 4 4 20 30 10 40 5 6 20 34 54 10 15 4 2 10 20 16 13
             

            Sample Output
            Case 1: 3 Case 2: 3 Case 3: -1

            // Bellman_Ford 
            #include <stdio.h>
            #include 
            <memory>
            #include 
            <iostream>
            #include 
            <algorithm>
            #include 
            <cstring>
            #include 
            <vector>
            #include 
            <map>
            #include 
            <cmath>
            #include 
            <set>
            #include 
            <queue>
            #include 
            <time.h> 
            #include 
            <limits>
            using namespace std;
            #define typev int
            #define N 1005
            #define inf 0x7fffffff 
            #define E (N*5) 
            const double pi = acos(-1.0); 
            struct e{
                
            int st, ed; 
                typev len;  
                
            void set(int _st, int _ed, typev _len){
                    st 
            = _st; 
                    ed 
            = _ed; 
                    len 
            = _len; 
                }
            }es[E];
            struct node{
                
            int h; 
                
            int i; 
            }nodes[N];
            e
            * fir[N]; 
            int n, en, d;
            int st, ed; 
            int vis[N], t[N], que[N]; 
            typev dist[N]; 
            inline 
            bool cmp(node n1, node n2){
                
            return n1.h < n2.h; 
            }
            inline 
            bool relax(int st, int ed, int len){
                
            if(dist[st] < inf && dist[ed] > dist[st] + len){
                    dist[ed] 
            = dist[st] + len;
                    
            return true
                }
                
            return false
            }
            bool Bellman_Ford(int n, int st){  //返回true表示有負(fù)環(huán)
                int i, j; 
                
            bool flag; 
                
            for(i = 0; i < n; i++) dist[i] = inf; 
                dist[st] 
            = 0
                
            for(i = 0; i < n; i++){
                    flag 
            = false
                    
            for(j = 0; j < en; j++){
                        
            if(relax(es[j].st, es[j].ed, es[j].len)) flag = true
                    }
                    
            if(!flag) break
                }
                
            return flag;  
            }
            int cnt = 0
            int main(){
            #ifndef ONLINE_JUDGE
                freopen(
            "in.txt""r", stdin); 
                
            //freopen("out.txt", "w", stdout); 
            #endif 

                
            int t, i, ans, u, v;
                scanf(
            "%d"&t);
                
            while(t--){
                    scanf(
            "%d%d"&n, &d);
                    
            for(i = 0; i < n; i++){
                        scanf(
            "%d"&nodes[i].h);
                        nodes[i].i 
            = i; 
                    }
                    en 
            = st = ed = 0
                    sort(nodes, nodes
            +n, cmp); 
                    st 
            = nodes[0].i; 
                    ed 
            = nodes[n - 1].i; 
                    
            if(st > ed) swap(st, ed); 
                    
            for(i = 0; i < n; i++) fir[i] = NULL; 
                    
            for(i = 1; i < n; i++){
                        es[en
            ++].set(i, i-1-1); 
                        u 
            = nodes[i].i; 
                        v 
            = nodes[i - 1].i; 
                        
            if(u < v){
                            es[en
            ++].set(u, v, d); 
                        }
            else es[en++].set(v, u, d); 
                    } 
                    
            if(!Bellman_Ford(n, st)) ans = dist[ed];
                    
            else ans = -1
                    printf(
            "Case %d: %d\n"++cnt, ans);

                }
                
            return 0;
            }



            posted on 2011-01-23 01:13 tw 閱讀(496) 評(píng)論(0)  編輯 收藏 引用 所屬分類: HDU題解

            <2025年6月>
            25262728293031
            1234567
            891011121314
            15161718192021
            22232425262728
            293012345

            導(dǎo)航

            統(tǒng)計(jì)

            常用鏈接

            留言簿

            文章分類

            文章檔案

            搜索

            最新評(píng)論

            久久亚洲精品中文字幕三区| 伊人热人久久中文字幕| 国产999精品久久久久久| 国产精品99久久99久久久| 亚洲色大成网站WWW久久九九| 亚洲va久久久久| 狠狠综合久久综合88亚洲| 亚洲伊人久久大香线蕉综合图片| 亚洲精品无码久久久久久| 国产精品99久久99久久久| 99精品伊人久久久大香线蕉| 蜜桃麻豆www久久国产精品| 精品国产99久久久久久麻豆| 国产Av激情久久无码天堂| 国产ww久久久久久久久久| 狠狠色丁香久久婷婷综合_中| 久久久久无码精品国产不卡| 久久天堂电影网| 国产精品久久久香蕉| 久久综合综合久久狠狠狠97色88| 亚洲美日韩Av中文字幕无码久久久妻妇 | 国产视频久久| 偷窥少妇久久久久久久久| 久久婷婷久久一区二区三区 | 精品久久人人做人人爽综合| 久久久久久久久久久久久久| 色综合久久综精品| av无码久久久久不卡免费网站| 日本精品一区二区久久久| 久久青草国产手机看片福利盒子| 东方aⅴ免费观看久久av| 久久久久亚洲AV无码专区网站| 国产精品久久久久jk制服| 国内精品九九久久精品| 亚洲国产成人久久一区WWW| 亚洲一本综合久久| 99久久精品国内| 久久精品国产久精国产思思| 欧美一区二区久久精品| 久久人人爽人人爽人人片AV麻豆 | 久久国产免费直播|