POJ 1738 An old Stone Game（石子合并）-白红宇

POJ 1738 An old Stone Game（石子合并）

阅读量：5806 次

发布时间：2019-06-18

本文共 2399 字，大约阅读时间需要 7 分钟。

题目链接：

题意：给出n堆石子，每次合并相邻的两堆石子，代价为这两堆石子的重量和，把所有石子合并为一堆，求最小代价。

思路：原始的DP方程为：f[i][j]=min(f[i][k]+f[k+1][j]+sum[j]-sum[i-1])。这个方法在这里是超时的。可以ac的解法是这样，道理目前不清楚：

（1）每次找到一个最小的i使得A[i-1]<=A[i+1]，将A[i-1]和A[i]合并为temp

（2）找到前面一个最大的j使得A[j]>temp，将temp放在j之后。

（3）重复1,2，直到剩余的堆数为1。

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               #define max(x,y) ((x)>(y)?(x):(y))#define min(x,y) ((x)<(y)?(x):(y))#define abs(x) ((x)>=0?(x):-(x))#define i64 long long#define u32 unsigned int#define u64 unsigned long long#define clr(x,y) memset(x,y,sizeof(x))#define CLR(x) x.clear()#define ph(x) push(x)#define pb(x) push_back(x)#define Len(x) x.length()#define SZ(x) x.size()#define PI acos(-1.0)#define sqr(x) ((x)*(x))#define MP(x,y) make_pair(x,y)#define EPS 1e-9#define FOR0(i,x) for(i=0;i
                
                 =0;i--)#define FORL1(i,a) for(i=a;i>=1;i--)#define FORL(i,a,b)for(i=a;i>=b;i--)#define rush() int CC;for(scanf("%d",&CC);CC--;)#define Rush(n) while(scanf("%d",&n)!=-1)using namespace std;void RD(int &x){scanf("%d",&x);}void RD(u32 &x){scanf("%u",&x);}void RD(i64 &x){scanf("%I64d",&x);}void RD(double &x){scanf("%lf",&x);}void RD(int &x,int &y){scanf("%d%d",&x,&y);}void RD(u32 &x,u32 &y){scanf("%u%u",&x,&y);}void RD(double &x,double &y){scanf("%lf%lf",&x,&y);}void RD(int &x,int &y,int &z){scanf("%d%d%d",&x,&y,&z);}void RD(int &x,int &y,int &z,int &t){scanf("%d%d%d%d",&x,&y,&z,&t);}void RD(u32 &x,u32 &y,u32 &z){scanf("%u%u%u",&x,&y,&z);}void RD(double &x,double &y,double &z){scanf("%lf%lf%lf",&x,&y,&z);}void RD(char &x){x=getchar();}void RD(char *s){scanf("%s",s);}void RD(string &s){cin>>s;}void PR(int x) {printf("%d\n",x);}void PR(int x,int y) {printf("%d %d\n",x,y);}void PR(int x,int y,int z) {printf("%d %d %d\n",x,y,z);}void PR(i64 x) {printf("%lld\n",x);}void PR(u32 x) {printf("%u\n",x);}void PR(double x) {printf("%.5lf\n",x);}void PR(char x) {printf("%c\n",x);}void PR(char *x) {printf("%s\n",x);}void PR(string x) {cout<
                 
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                    =2&&A[j]>=A[j-2]) temp=T-j,combine(j-1),j=T-temp;}int main(){ Rush(n) { if(!n) break; int i; FOR0(i,n) RD(A[i]); T=1;ans=0; FOR1(i,n-1) { A[T++]=A[i]; while(T>=3&&A[T-3]<=A[T-1]) combine(T-2); } while(T>1) combine(T-1); PR(ans); } return 0;}

转载于:https://www.cnblogs.com/jianglangcaijin/archive/2013/04/24/3040815.html

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