uva507 - Jill Rides Again(JIll又骑车了)

思路不难，就是暴力，

两层暴力。不过要剪枝，

考虑那次循环是无用的，则跳过那样的耗时、

假如1~m目前求得的最大值是M。

那么如果a[1]>0的话，就没有必要让i=2再来一次循环了，因为a[1]>0,所以2~m最大值会比M少a[1].，所以把这样的无用功剪去就好了

代码如下：

#include <cstdio>
#define M 20010
int n, a[M];
int main ()
{
    int cas, t = 0, n, sum, maxi, maxj, max, m;
    scanf("%d",&cas);
    while(t++<cas)
    {
        scanf("%d",&n);
        for(int i = 1; i < n; i++) scanf("%d",&a[i]);
        a[0] = -1; sum = -1; maxi = maxj = max = m = 0;
        for(int i = 1; i < n; i++)
        {
            m+=a[i-1];//剪枝条件判断
            if(m>=0) continue;
            sum = 0;
            for(int j = i; j < n; j++)
            {
                sum+=a[j];//printf("%d  %d = %d\n",i,j,sum);
                if(sum>max) {max = sum; maxi = i; maxj = j;}
                if(sum==max&&maxj-maxi<j-i) {max = sum; maxi = i; maxj = j;}
            }
            m = 0;
        }
        if(max<=0) printf("Route %d has no nice parts\n",t);
        else printf("The nicest part of route %d is between stops %d and %d\n",t,maxi, maxj+1);
    }
    return 0;
}