Codeforces Beta Round #6 (Div. 2 Only) D. Lizards and Basements 2
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题目大意
有排成一排的
解题思路
这是一个动态规划问题。定义一个四维
题目代码
#include <set>#include <map>#include <queue>#include <math.h>#include <vector>#include <string>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <cctype>#include <algorithm>#include <time.h>#define eps 1e-10#define pi acos(-1.0)#define inf 107374182#define inf64 1152921504606846976#define lc l,m,tr<<1#define rc m + 1,r,tr<<1|1#define zero(a) fabs(a)<eps#define iabs(x) ((x) > 0 ? (x) : -(x))#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (min(SIZE,sizeof(A))))#define clearall(A, X) memset(A, X, sizeof(A))#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))#define memcopyall(A, X) memcpy(A , X ,sizeof(X))#define max( x, y ) ( ((x) > (y)) ? (x) : (y) )#define min( x, y ) ( ((x) < (y)) ? (x) : (y) )using namespace std;int n,a,b,h[15];int dp[15][20][20][20];struct node{ int i,j,k,l;} p[15][20][20][20],temp;int main(){ scanf("%d%d%d",&n,&a,&b); for(int i=0; i<n; i++) { scanf("%d",&h[i]); h[i]++; } clearall(dp,0x3f); dp[1][h[0]][h[1]][h[2]]=0; int x,y,z,d1,d2,d3,init = dp[0][0][0][0]; //cout<<init; for(int i=1; i<n-1; i++) { for(int j=h[i-1]; j>=0; j--) { for(int k=h[i]; k>=0; k--) { for(int l=h[i+1]; l>=0; l--) { if(dp[i][j][k][l]!=init) { d1=max(0,j-b); d2=max(0,k-a); d3=max(0,l-b); if(dp[i][d1][d2][d3]>dp[i][j][k][l]+1) { dp[i][d1][d2][d3]=dp[i][j][k][l]+1; p[i][d1][d2][d3].j=j; p[i][d1][d2][d3].k=k; p[i][d1][d2][d3].l=l; p[i][d1][d2][d3].i=i; } if(d1==0&&dp[i+1][d2][d3][h[i+2]]>dp[i][j][k][l]+1) { dp[i+1][d2][d3][h[i+2]]=dp[i][j][k][l]+1; p[i+1][d2][d3][h[i+2]].i=i; p[i+1][d2][d3][h[i+2]].j=j; p[i+1][d2][d3][h[i+2]].k=k; p[i+1][d2][d3][h[i+2]].l=l; } if(j==0&&dp[i+1][k][l][h[i+2]]>dp[i][j][k][l]) { dp[i+1][k][l][h[i+2]]=dp[i][j][k][l]; p[i+1][k][l][h[i+2]]=p[i][j][k][l]; } } } } } } printf("%d\n",dp[n-1][0][0][0]); temp.i=n-1; temp.j=0; temp.k=0; temp.l=0; while(!(temp.i==1&&temp.j==h[0]&&temp.k==h[1]&&temp.l==h[2])) { temp = p[temp.i][temp.j][temp.k][temp.l]; printf("%d ",temp.i+1); } return 0;}
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