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题意:给n(n <= 800,000)头牛，每头牛都有一个高度h,每头牛都只能看到右边比它矮的牛的头发，将每头牛看到的牛的头发加起来为多少?
思路:每头要进栈的牛，将栈顶比其矮的牛出栈，因为这些牛都没有机会看到更后面的牛了，所以出栈;这时加上栈中的元素个数即可；
#include<iostream> #include<cstdio> #include<cstring> #include<string.h> #include<algorithm> #include<vector> #include<cmath> #include<stdlib.h> #include<time.h> #include<stack> #include<set> #include<map> #include<queue> using namespace std; #define rep0(i,l,r) for(int i = (l);i < (r);i++) #define rep1(i,l,r) for(int i = (l);i <= (r);i++) #define rep_0(i,r,l) for(int i = (r);i > (l);i--) #define rep_1(i,r,l) for(int i = (r);i >= (l);i--) #define MS0(a) memset(a,0,sizeof(a)) #define MS1(a) memset(a,-1,sizeof(a)) #define MSi(a) memset(a,0x3f,sizeof(a)) #define inf 0x3f3f3f3f #define lson l, m, rt << 1 #define rson m+1, r, rt << 1|1 typedef pair<int,int> PII; #define A first #define B second #define MK make_pair typedef __int64 ll; template<typename T> void read1(T &m) { T x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} m = x*f; } template<typename T> void read2(T &a,T &b){read1(a);read1(b);} template<typename T> void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);} template<typename T> void out(T a) { if(a>9) out(a/10); putchar(a%10+‘0‘); } const int MAXN = 8e4+7; int stk[MAXN]; int main() { ll p = 0,ans = 0,n,h; read1(n); rep0(i,0,n){ read1(h); while(p && h >= stk[p]) p--; stk[++p] = h; ans += p-1; //cout<<p<<" .. \n"; } printf("%I64d\n",ans); return 0; }
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原文地址：http://www.cnblogs.com/hxer/p/5294165.html
