UVA1347Tour

这道题，关键在于，告诉你，递归的效率可以很低！

先给出我的递归程序

//UVA1347Tour
#include
#include 
#include
#include
using namespace std;

const int MAXN = 1e3 + 10;
int n;
double x[MAXN], y[MAXN];
double d[MAXN][MAXN];
double dis[MAXN][MAXN];
double dp(int i, int j) {
	int tmp = 0;
	if(i < j) {
		tmp = i; i = j; j = tmp;
	} 
	if(i == n - 1) { d[i][j] = d[j][i] = dis[n - 1][n] + dis[n][j];
	 //printf("d = %lf\n", d[i][j]);
	 return d[i][j]; }
	return d[i][j] = d[j][i] = min(dp(i + 1, j) + dis[i + 1][i], dp(i + 1, i) + dis[j][i + 1]);
	
}
int main() {
	freopen("UVA1347out.txt", "w", stdout);
	while(scanf("%d", &n) == 1) {
		memset(dis, 0, sizeof(dis));
		memset(d, 0, sizeof(d));
		for(int i = 1; i <= n; i++) scanf("%lf%lf", &x[i], &y[i]);
		for(int i = 1; i <= n; i++) 
		    for(int j = i + 1; j <= n; j++)
		    	if(dis[i][j] != 0) continue;
		    	else dis[i][j] = dis[j][i] = sqrt((x[j] - x[i]) * (x[j] - x[i]) + (y[j] - y[i]) * (y[j] - y[i]));
		    	//计算两点间欧氏距离
		//for(int i - ) 
		printf("%.2lf\n", dp(2, 1) + dis[2][1]);
		/*for(int i = 1; i <= n; i++) {
			for(int j = i + 1; j <= n; j++) {
				printf("%4lf ", dis[j][i]);
			}
			printf("\n");
		}*/
	}
}
/*
3
1 1
2 3
3 1
4
1 1
2 3
3 1
4 2
*/

虽然udebug的数据过了，但交上去3s都没跑完，直接T

所以要强行改成循环

下面是AC程序

//UVA1347Tour
#include
#include 
#include
#include
using namespace std;

const int MAXN = 1e3 + 10;
int n;
double x[MAXN], y[MAXN];
double d[MAXN][MAXN];
double dis[MAXN][MAXN];
int main() {
	//freopen("UVA1347out.txt", "w", stdout);
	while(scanf("%d", &n) == 1) {
		memset(dis, 0, sizeof(dis));
		memset(d, 0, sizeof(d));
		for(int i = 1; i <= n; i++) scanf("%lf%lf", &x[i], &y[i]);
		for(int i = 1; i <= n; i++) 
		    for(int j = i + 1; j <= n; j++)
		    	if(dis[i][j] != 0) continue;
		    	else dis[i][j] = dis[j][i] = sqrt((x[j] - x[i]) * (x[j] - x[i]) + (y[j] - y[i]) * (y[j] - y[i]));
		    	//计算两点间欧氏距离
		for(int i = n - 1; i >= 2; i--) {
			for(int j = 1; j < i; j++) {
				if(i == n - 1) d[i][j] = dis[i][n] + dis[j][n];
				else d[i][j] = min(d[i + 1][j] + dis[i + 1][i], d[i + 1][i] + dis[j][i + 1]);
			}
		}
		printf("%.2lf\n", d[2][1] + dis[2][1]);
		/*for(int i = 1; i <= n; i++) {
			for(int j = i + 1; j <= n; j++) {
				printf("%4lf ", dis[j][i]);
			}
			printf("\n");
		}*/
	}
}
/*
3
1 1
2 3
3 1
4
1 1
2 3
3 1
4 2
*/