C 解法, 执行用时: 2ms, 内存消耗: 232KB, 提交时间: 2019-01-22

#include <stdio.h>
#include <string.h>
int main(void){
    int T,flag;
    long n;
    long i,k,min,y;
    int mincost;
    char c[4] = {'E','S','W','N'};
    char d[100000];
    long a[4][100000],b[100000];
    scanf("%d",&T);
    for(y = 0;y < T; y++){
        mincost = 0;
        //memset(d, 0x00, sizeof (char) * 100000);
        //memset(a,0,sizeof(a));
        //memset(b,0,sizeof(b));
        scanf("%ld",&n);
            for(i = 0;i < 4;i++){
                for(k = 0;k < n;k++){
                    scanf("%ld",&a[i][k]);
                }
            }
        for(i = 0;i < n;i++){
            min = a[0][i];
            flag = 0;
            for(k = 1;k < 4;k++){
                if(min > a[k][i]){
                    min = a[k][i];
                    flag = k;
                }
            }
            mincost = mincost + min;
            d[i] = c[flag];
        }
        printf("%d\n",mincost);
        for (i = 0;i < n;i++){
            printf("%c",d[i]);
        }
        printf("\n");
    }
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 340KB, 提交时间: 2019-01-28

#include <stdio.h>
int t, k;
int e[100005], s[100005], w[100005], n[100005];
char dir[100005];
void init()
{
	scanf("%d", &t);
}
int min_of_four(int a, int b, int c, int d, char* ch) //min value of four direction's power
{
	int min = a;
	*ch = 'E';
	if(b < min)
	{
		min = b;
		*ch = 'S';
	}
	if(c < min)
	{
		min = c;
		*ch = 'W';
	}
	if(d < min)
	{
		min = d;
		*ch = 'N';
	}
	return min;
}
void fun()
{
	for(int i = 0; i < t; i++)
	{
		scanf("%d", &k);
		for(int j = 0; j < k; j++)
			scanf("%d", &e[j]);
		for(int j = 0; j < k; j++)
			scanf("%d", &s[j]);
		for(int j = 0; j < k; j++)
			scanf("%d", &w[j]);
		for(int j = 0; j < k; j++)
			scanf("%d", &n[j]);
		int cnt = 0;
		for(int i = 0; i < k; i++)
			cnt += min_of_four(e[i], s[i], w[i], n[i], &dir[i]);
		printf("%d\n", cnt);
		for(int i = 0; i < k; i++)
			printf("%c", dir[i]);
		printf("\n");
	}
}

int main()
{
	init();
	fun();
	return 0;
}