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                   uva 10564 Paths through the Hourglass

Paths through the Hourglass
Input: Standard Input

Output: Standard Output

Time Limit: 2 Seconds

A path is described with an integer representing the starting point in the first row (the leftmost cell being 0) followed by a direction string containing the letters L and R, telling whether to go to the left or right. For instance, the path to the right is described as 2 RRRLLRRRLR.

Given the values of each cell in an hourglass as well as an integer S, calculate the number of distinct paths with value S. If at least one path exist, you should also print the path with the lowest starting point. If several such paths exist, select the one which has the lexicographically smallest direction string.

Input

The input contains several cases. Each case starts with a line containing two integers N and S (2≤N≤20, 0≤S<500), the number of cells in the first row of the hourglass and the desired sum. Next follows 2N-1 lines describing each row in the hourglass. Each line contains a space separated list of integers between 0 and 9 inclusive. The first of these lines will contain N integers, then N-1, ..., 2, 1, 2, ..., N-1, N.

The input will terminate with N=S=0. This case should not be processed. There will be less than 30 cases in the input.

Output

For each case, first output the number of distinct paths. If at least one path exist, output on the next line the description of the path mentioned above. If no path exist, output a blank line instead.

Sample Input                             Output for Sample Input

题目大意：给出一个用数字组成的沙漏。要求求出从第一层任意一点开始，往下走，直到最后一层，走过的路径和等于目标值的方案数。输出方案数，并输出字典序最小的方案的具体走法，若方案数为0，则输出一个空行。

解题思路：

1）如果直接读入的话，dp时判断左右会稍微有点麻烦，因为上下会相反。所以读入的时候可以稍作处理，在方向判断上就不会容易出错。

2）dp[i][j][k]数组记录的是格子(i, j)加到最后一行和为k的方案数。

3）dp数组和最后记录总方案数的ans要用long long。

4）dp从最后一行开始，先将最后一行全部置为1。

5）状态转移方程：dp[i][j][k] = dp[i + 1][j][k - v] + dp[i + 1][j + 1][k - v];   v 为（i， j）格子的值。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define N 55
#define M 505
typedef long long ll;
using namespace std;
int n, m;
ll ans, dp[N][N][M];
int num[N * 2][N * 2];
void DP() { //dp[i][j][k]数组记录的是格子(i, j)加到最后一行为k的方案数
	for (int i = 2 * n - 2; i >= n; i--) {
		for (int j = i; j >= n - 1; j--) {
			if (i == 2 * n - 2) {
				dp[i][j][num[i][j]] = 1;
			}
			else {
				int w, v;
				w = v = num[i][j];
				for (; v <= m; v++) {
					dp[i][j][v] = dp[i + 1][j][v - w] + dp[i + 1][j + 1][v - w];
				}
			}
		}
	}
	int cnt = n - 1;
	for (int i = n - 1; i >= 0; i--) {
		for (int j = cnt; j <= n - 1; j++) {
			int w, v;
			w = v = num[i][j];
			for (; v <= m; v++) {
				dp[i][j][v] = dp[i + 1][j][v - w] + dp[i + 1][j + 1][v - w];
			}
			if (i == 0) {
				ans += dp[i][j][m];
			}
		}
		cnt--;
	}
}
void outPut() {
	printf("%lld\n", ans);
	int flag = 0;
	for (int i = 0; i < n; i++) {
		if (dp[0][i][m]) {
			flag = 1;
			printf("%d", i);
			int rec = i, s = m - num[0][i], flag2 = 1;
			for (int j = 1; j < 2 * n - 1; j++) {
				if (flag2) {
					printf(" ");
					flag2 = 0;
				}
				if (dp[j][rec][s]) {
					printf("L");
					s -= num[j][rec];
				}
				else if (dp[j][rec + 1][s]){
					printf("R");
					rec++;
					s -= num[j][rec];
				}
			}
			break;
		}
	}
	if (!flag) {
		printf("\n");
	}
	else printf("\n");
}
int main() {
	while (scanf("%d %d", &n, &m) == 2) {
		if (n == 0 && m == 0) break;
		memset(dp, 0, sizeof(dp));
		memset(num, -1, sizeof(num));
		for (int i = 0; i < n; i++) {
			for (int j = i; j < n; j++) {
				scanf("%d", &num[i][j]);
			}
		}	
		for (int i = n; i < 2 * n - 1; i++) {
			for (int j = 0; j < 2 + i - n; j++) {
				scanf("%d", &num[i][j + n - 1]);
			}
		}
		ans = 0;
		DP();
		outPut();
	}
	return 0;
}

uva 10564 Paths through the Hourglass （DP）

标签：c语言   acm   uva   dp   

原文地址：http://blog.csdn.net/llx523113241/article/details/44245471