uva 10285 Longest Run on a Snowboard (DP)

Longest Run on a SnowboardInput: standard inputOutput: standard outputTime Limit: 5 secondsMemory Limit:32 MBMichael likes snowboarding. That's not very surprising, since snowboarding is really great. The bad thing is that in order to gain speed, the area must slide downwards. Another disadvantage is that when you've reached the bottomof the hill you have to walk up again or wait for the ski-lift.Michael would like to know how long the longest run in an area is. That area is given by a grid of numbers, defining the heights at those points. Look at this example:

�1� 2� 3� 4 5

16 17 18 19 6

15 24 25 20 7

14 23 22 21 8

13 12 11 10 9

One can slide down from one point to a connected other one if and only if the height decreases. One point is connected to another if it's at left, at right, above or below it. In the sample map, a possible slide would be24-17-16-1 (start at 24, end at 1). Of course if you would go25-24-23-...-3-2-1, it would be a much longer run. In fact, it's the longest possible.InputThe first line contains the number of test cases N. Each test case starts with a line containing the name (it's a single string), the number of rowsR and the number of columns C. After that followR lines with C numbers each, defining the heights.R and C won't be bigger than 100, N not bigger than15 and the heights are always in the range from 0 to100.For each test case, print a line containing the name of the area, a colon, a space and the length of the longest run one can slide down in that area.

Sample Input

2

Feldberg 10 5

56 14 51 58 88

26 94 24 39 41

24 16 8 51 51

76 72 77 43 10

38 50 59 84 81

5 23 37 71 77

96 10 93 53 82

94 15 96 69 9

74 0 62 38 96

37 54 55 82 38

Spiral 5 5

1 2 3 4 5

16 17 18 19 6

15 24 25 20 7

14 23 22 21 8

13 12 11 10 9

Sample Output

Feldberg: 7

Spiral: 25

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>using namespace std;const int maxn = 110;const int dr[4] = {0 , 0 , -1 , 1};const int dl[4] = {1 , -1 , 0 , 0};int N , n , m;int h[maxn][maxn] , dp[maxn][maxn];struct Moutain{int x , y , h;Moutain(int a = 0 , int b = 0 , int c = 0){x = a, y = b , h = c;}};vector<Moutain> M;int ans;string Name;bool cmp(Moutain m1 , Moutain m2){return m1.h>m2.h;}void initial(){for(int i = 0;i < maxn;i++){for(int j = 0;j < maxn;j++){h[i][j] = maxn;dp[i][j] = 0;}}ans = 0;M.clear();}void readcase(){cin >> Name >> n >> m;for(int i = 1;i <= n;i++){for(int j =1;j <= m;j++){cin >> h[i][j];M.push_back(Moutain(i , j , h[i][j]));}}}void computing(){sort(M.begin() , M.end() , cmp);for(int i = 0;i < M.size();i++){for(int k = 0;k < 4;k++){if(h[M[i].x+dr[k]][M[i].y+dl[k]] < h[M[i].x][M[i].y]){dp[M[i].x+dr[k]][M[i].y+dl[k]] = max(dp[M[i].x+dr[k]][M[i].y+dl[k]] , dp[M[i].x][M[i].y]+1);ans = max(ans , dp[M[i].x+dr[k]][M[i].y+dl[k]]);}}}cout << Name << ": " << ans+1 << endl;}int main(){int t;cin >> t;while(t--){initial();readcase();computing();}return 0;}

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