UVA 10285 Longest Run on a Snowboard

题目大意：求矩阵中最长递减路径的长度。
解析：有两种解法，一种是dfs，还有一种是dp，因为最大的矩阵为100*100，所以直接暴力dfs就可以过。
dp解法：先把点的坐标和数值存于结构体中，然后将结构体从小到大排序，然后按照数值从小到大dp
如果当前点周围的高度h[newx][newy] > h[x][y]，且dp[x][y] >= dp[newx][newy]

那么dp[newx][newy] = dp[x][y] + 1;

dfs解法：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 105;
const int dx[] = {-1,0,1,0};
const int dy[] = {0,-1,0,1};
int h[N][N];
int r,c;
int maxh;
void dfs(int x,int y,int sum) {
	int newx , newy;
	if(sum > maxh) {
		maxh = sum;
	}
	for(int i = 0; i < 4; i++) {
		newx = x + dx[i];
		newy = y + dy[i];
		if(newx < 0 || newx >= r || newy < 0 || newy >= c) {
			continue;
		}
		if(h[newx][newy] < h[x][y]) {
			dfs(newx,newy,sum+1);
		}
	}
}
int main() {
	int T;
	char name[N];
	scanf("%d",&T);
	while(T--) {
		scanf("%s%d%d",name,&r,&c);
		for(int i = 0; i < r; i++) {
			for(int j = 0; j < c; j++) {
				scanf("%d",&h[i][j]);
			}
		}
		maxh = -INF;
		for(int i = 0; i < r; i++) {
			for(int j = 0; j < c; j++) {
				dfs(i,j,0);
			}
		}
		printf("%s: %d\n",name,maxh+1);
	}
	return 0;
}

dp解法：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 105;
const int dx[] = {-1,0,1,0};
const int dy[] = {0,-1,0,1};
struct Point {
	int x,y,h;
	Point() {}
	Point(int _x,int _y,int _h) {
		x = _x;
		y = _y;
		h = _h;
	}
}p[N*N];
int h[N][N];
int dp[N][N];
int r,c;
bool cmp(Point a,Point b) {
	return a.h < b.h;
}
void update(int x,int y) {
	int newx,newy;
	int tmp[4];
	for(int i = 0; i < 4; i++) {
		newx = x + dx[i];
		newy = y + dy[i];
		if(newx < 0 || newx >= r || newy < 0 || newy >= c) {
			continue;
		}
		if(h[newx][newy] > h[x][y] && dp[x][y] >= dp[newx][newy]) {
			dp[newx][newy] = dp[x][y] + 1;
		}
	}
}
int main() {
	int T;
	char name[N];
	scanf("%d",&T);
	while(T--) {
		scanf("%s%d%d",name,&r,&c);
		int tot = 0;
		for(int i = 0; i < r; i++) {
			for(int j = 0; j < c; j++) {
				scanf("%d",&h[i][j]);
				p[tot++] = Point(i,j,h[i][j]); 
			}
		}
		sort(p,p+tot,cmp);
		memset(dp,0,sizeof(dp));
		for(int i = 0; i < tot; i++) {
			update(p[i].x,p[i].y);
		}
		int maxh = -INF;
		for(int i = 0; i < r; i++) {
			for(int j = 0; j < c; j++) {
				maxh = max(maxh,dp[i][j]);
			}
		}
		printf("%s: %d\n",name,maxh+1);
	}
	return 0;
}