UVA10564-----Paths through the Hourglass-----简单的计数DP
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本文出自:http://blog.csdn.net/dr5459
题目地址:
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1505
题目意思:
给你2*n-1行
第一行有n个,第n行有1个,然后第2*n-1行有n个,一个沙漏状
里面每个单元都有一个数字
给你S,问你从顶走到底,和为S有多少种,并输出其中字典序最小的路径
解题思路:
设dp[i][j][k] 表示从下往上的满足条件的i行j列和为K的种数,这里要用long long
令tag数组表示是往左还是往右
简单的递推一下就可以出来,另外要注意这里是从0开始编号的
下面上代码:
#include<cstdio>#include<iostream>#include<cstring>using namespace std;#define LL long longconst int maxn = 50;const int maxs = 550;LL dp[maxn][maxn][maxs];bool tag[maxn][maxn][maxs][2];int num[maxn][maxn];int n,s;int main(){ while(~scanf("%d%d",&n,&s) && n+s) { for(int i=1;i<=n;i++) { for(int j=1;j<=n-i+1;j++) scanf("%d",&num[i][j]); } for(int i=n+1;i<=2*n-1;i++) { for(int j=1;j<=i-n+1;j++) scanf("%d",&num[i][j]); } memset(dp,0,sizeof(dp)); memset(tag,false,sizeof(tag)); //初始化 for(int i=1;i<=n;i++) { int tmp = s-num[2*n-1][i]; dp[2*n-1][i][tmp] = 1; } //从下往上算 for(int i=2*n-2;i>=n;i--) { for(int j=1;j<=i-n+1;j++) { for(int k=0;k<=s;k++) { int tmp = k+num[i][j]; if(dp[i+1][j][tmp] != 0 ) { dp[i][j][k] += dp[i+1][j][tmp]; tag[i][j][k][0] = true; } if(dp[i+1][j+1][tmp] != 0) { dp[i][j][k] += dp[i+1][j+1][tmp]; tag[i][j][k][1] = true; } } } } for(int i=n-1;i>=1;i--) { for(int j=1;j<=n-i+1;j++) { for(int k=0;k<=s;k++) { int tmp = k+num[i][j]; if(dp[i+1][j-1][tmp] != 0 ) { dp[i][j][k] += dp[i+1][j-1][tmp]; tag[i][j][k][0] = true; } if(dp[i+1][j][tmp] != 0) { dp[i][j][k] += dp[i+1][j][tmp]; tag[i][j][k][1] = true; } } } } int ansi = -1; LL ans = 0; for(int i=1;i<=n;i++) { if(dp[1][i][0] !=0 && ansi == -1) ansi = i; ans += dp[1][i][0]; } cout<<ans<<endl; if(ans == 0) { cout<<endl; continue; } cout<<ansi-1<<" "; //从0开始编号 int j = ansi; int now = 0; for(int i=1;i<=2*n-1;i++) { int pos = j; if(tag[i][pos][now][0]) { cout<<"L"; if(i<n) j--; } else if(tag[i][pos][now][1]) { cout<<"R"; if(i>=n) j++; } now += num[i][pos]; } cout<<endl; } return 0;}
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