题目链接:
http://poj.org/problem?id=2796
题意:
给你一个序列，让你求出一段区间，（在这段区间之内的最小值*这段区间所有元素之和）的最大值……
题解:
http://www.cnblogs.com/ziyi–caolu/archive/2013/06/23/3151556.html
代码:
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <stack> 5 using namespace std; 6 typedef long long ll; 7 #define MS(a) memset(a,0,sizeof(a)) 8 #define MP make_pair 9 #define PB push_back 10 const int INF = 0x3f3f3f3f; 11 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL; 12 inline ll read(){ 13 ll x=0,f=1;char ch=getchar(); 14 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 15 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 16 return x*f; 17 } 18 ////////////////////////////////////////////////////////////////////////// 19 const int maxn = 1e5+10; 20 21 struct node{ 22 ll num,pre,nxt; 23 ll pos; 24 }; 25 26 ll a[maxn], sum[maxn]; 27 28 int main(){ 29 int n; 30 while(scanf("%d",&n)==1){ 31 stack<node> s; 32 sum[0] = 0; 33 for(int i=1; i<=n; i++){ 34 a[i] = read(); 35 sum[i] = sum[i-1]+a[i]; 36 } 37 node k; 38 k.num = a[1]; k.pre = k.nxt = 1; k.pos = 1; 39 s.push(k); 40 ll ans = -1,x,y; 41 for(int i=2; i<=n; i++){ 42 node t; 43 t.num = a[i]; t.pre = t.nxt = 1; t.pos = i; 44 while(!s.empty() && t.num<=s.top().num){ 45 node tt = s.top(); 46 s.pop(); 47 if(!s.empty()) 48 s.top().nxt += tt.nxt; 49 t.pre += tt.pre; 50 ll tmp = tt.num*(sum[tt.pos+tt.nxt-1]-sum[tt.pos-tt.pre]); 51 // cout << tt.pos << " " << tt.pre << " " << tt.nxt << " " << tmp << endl; 52 if(ans < tmp){ 53 ans = tmp; 54 x = tt.pos-tt.pre+1; 55 y = tt.pos+tt.nxt-1; 56 } 57 } 58 s.push(t); 59 } 60 61 while(!s.empty()){ 62 node tt = s.top(); 63 s.pop(); 64 if(!s.empty()) 65 s.top().nxt += tt.nxt; 66 67 ll tmp = tt.num*(sum[tt.pos+tt.nxt-1]-sum[tt.pos-tt.pre]); 68 if(ans < tmp){ 69 ans = tmp; 70 x = tt.pos-tt.pre+1; 71 y = tt.pos+tt.nxt-1; 72 } 73 } 74 75 cout << ans << endl << x << " " << y << endl; 76 } 77 78 return 0; 79 } 80 // http://www.cnblogs.com/ziyi--caolu/archive/2013/06/23/3151556.html