10285 Longest Run on a Snowboard Michael 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 bottom of 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 be 24-17-16-1 (start at 24, end at 1). Of course if you would go 25-24-23-…-3-2-1, it would be a much longer run. In fact, it’s the longest possible. Input The 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 rows R and the number of columns C. After that follow R lines with C numbers each, defining the heights. R and C won’t be bigger than 100, N not bigger than 15 and the heights are always in the range from 0 to 100. Output 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 Universidad de Valladolid OJ: 10285 – Longest Run on a Snowboard 2/2 14 23 22 21 8 13 12 11 10 9 Sample Output Feldberg: 7 Spiral: 25
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/*Solve:*/;maxn=105;const int inf=999999999;LL;void Max(int&a,int b){if(a<b)a=b;}char name[50];int a[maxn][maxn];int dp[maxn][maxn];//end with (i,j) you can get the max lengthint n,m;int dx[]={0,0,1,-1};int dy[]={1,-1,0,0};int dfs(int x,int y){//表示从(x,y)这个点出发，所能到达的最大长度if(dp[x][y]){return dp[x][y];}int ans=0;for(int i=0;i<4;i++){int xx=dx[i]+x;int yy=dy[i]+y;if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&a[xx][yy]<a[x][y]){ans=max(ans,dfs(xx,yy));//记录下这个点出发的四个方向中最大的那个}}dp[x][y]+=ans+1;//最大值加上当前这个点return dp[x][y];}int main(){int T;scanf(“%d”,&T);while(T–){scanf(“%s%d%d”,name,&n,&m);int mxv=0,pi,pj;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){scanf(“%d”,&a[i][j]);if(a[i][j]>mxv){mxv=a[i][j];pi=i;pj=j;}}}cl(dp,0);int mx=1;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){mx=max(mx,dfs(i,j));//遍历每一个点}}printf(“%s: %d\n”,name,mx);}return 0;}/*2Feldberg 10 556 14 51 58 8826 94 24 39 4124 16 8 51 5176 72 77 43 1038 50 59 84 815 23 37 71 7796 10 93 53 8294 15 96 69 974 0 62 38 9637 54 55 82 38Spiral 5 51 2 3 4 516 17 18 19 615 24 25 20 714 23 22 21 813 12 11 10 9*/
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