UVA 11054 Wine trading in Gergovia

我的思路刚开始就是从i开始找一直左右最近的需要酒的居民然后和他交易,很不幸的TLE了.

后来想了个方法用pre数组表示上一个需要酒的居民,nex表示下一个需要酒的居民,然后每次先按照pre找左边最近需要酒的居民和他交易,更新pre,如果左边没有居民需要了那就往右边找,更新nex,直到卖完酒.更新的时候如果v[i]==0表示这个居民不需要酒了,就while循环往上更新pre直到找到一个需要酒的居民或者碰到边界,nex同理.

用了2.766s,擦过..

#include <iostream>#include <cstdio>using namespace std;const int maxn = 100010;int v[maxn], pre[maxn], nex[maxn], n;void readInt(int & x){x = 0;char ch = getchar();int sign = 1;while(!(ch >= '0' && ch <= '9')){if(ch == '-'){sign = -1;}ch = getchar();}while (ch >= '0' && ch <= '9'){x = x * 10 + (ch - '0');ch =getchar();}x *= sign;}int main(){while (scanf("%d", &n) && n){int prei = -1, nexi = n;long long ans = 0;for (int i = 0; i < n; ++i){readInt(v[i]);pre[i] = prei;if(v[i] > 0){prei = i;}}for (int i = n - 1; i >= 0; --i){nex[i] = nexi;if(v[i] > 0){nexi = i;}}for (int i = 0; i < n; ++i){if(v[i] < 0){int l = pre[i], r = nex[i];while (v[i]){while(l != -1 && !v[l]){l = pre[l];}if(l != -1){int t = -v[i];int val = min(t, v[l]);v[i] += val;v[l] -= val;while(v[l] == 0 && pre[l] !=  -1 && !v[pre[l]]){//压缩路径上去pre[l] = pre[pre[l]];}ans += val * (i - l);}else{while (r != n && !v[r]){r = nex[r];}int t = -v[i];int val = min(t, v[r]);v[i] += val;v[r] -= val;while(v[r] == 0 && nex[r] !=  n && !v[nex[r]]){nex[r] = nex[nex[r]];}ans += val * (r - i);}}}}printf("%lld\n", ans);}return 0;}

后来找了下报告发现有扫描的方法 很巧妙.

想象下有一个人从0开始到n - 1一路走过去,见到卖酒的人收下来,见到买酒的就把身上的酒卖给他(如果身上没酒了就相当于负的,欠着).

答案的就是每个位置当前身上的携带或者欠的酒数量的和.

#include <iostream>#include <cstdio>#include <cmath>using namespace std;const int maxn = 100010;int V[maxn], n;void readInt(int & x){x = 0;char ch = getchar();int sign = 1;while(!(ch >= '0' && ch <= '9')){if(ch == '-'){sign = -1;}ch = getchar();}while (ch >= '0' && ch <= '9'){x = x * 10 + (ch - '0');ch =getchar();}x *= sign;}int main(){while (scanf("%d", &n) && n){long long ans = 0;int curWines = 0;for (int i = 0; i < n; ++i){readInt(V[i]);}for (int i = 0; i < n; ++i){ans += abs(curWines);curWines += V[i];}printf("%lld\n", ans);}}