uva507 - Jill Rides Again(JIll又骑车了)

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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 20010int 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;}