UVa 10564 Paths through the Hourglass (DP)

题意：从最上面走到最下面，使得路过的数求和为s，并输出编号最小的一组路径。

析：基本动规，dp[i][j][s] 从最下面到 i，j 和为s，路径数，要么从左面要么从右，求和就好了，注意上面和下面的不太一样，要分别求解。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 150 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL dp[45][25][505];
int a[45][25];

void print(int d, int idx, int ans){
    if(d + 1 == 2 * n)  return ;
    if(d < n){
        if(1 == d)  for(int i = 1; i <= n; ++i)if(dp[d][i][ans]){
            printf("%d ", i-1);
            if(dp[d+1][i-1][ans-a[d][i]]){
                putchar('L');
                print(d+1, i-1, ans - a[d][i]);
                return ;
            }
            else{
                putchar('R');
                print(d+1, i, ans - a[d][i]);
                return ;
            }
        }
        if(dp[d+1][idx-1][ans-a[d][idx]]){
                putchar('L');
                print(d+1, idx-1, ans - a[d][idx]);
                return ;
        }
        else{
                putchar('R');
                print(d+1, idx, ans - a[d][idx]);
                return ;
        }
    }
    else{
        if(dp[d+1][idx][ans-a[d][idx]]){
                putchar('L');
                print(d+1, idx, ans - a[d][idx]);
                return ;
        }
        else{
                putchar('R');
                print(d+1, idx+1, ans - a[d][idx]);
                return ;
        }
    }
}

int main(){
    while(scanf("%d %d", &n, &m) == 2 && m + n){
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n-i+1; ++j)
                scanf("%d", &a[i][j]);
        for(int i = n+1; i < 2*n; ++i)
            for(int j = 1; j <= i-n+1; ++j)
                scanf("%d", &a[i][j]);

        memset(dp, 0, sizeof dp);
        for(int i = 1; i <= n; ++i)  dp[2*n-1][i][a[2*n-1][i]] = 1;
        for(int i = 2*n-2; i >= n; --i)
            for(int j = 1; j <= i-n+1; ++j)
                for(int k = a[i][j]; k <= m; ++k)
                    dp[i][j][k] = dp[i+1][j][k-a[i][j]] + dp[i+1][j+1][k-a[i][j]];
        for(int i = n-1; i > 0; --i)
            for(int j = 1; j <= n-i+1; ++j)
                for(int k = a[i][j]; k <= m; ++k)
                    dp[i][j][k] = dp[i+1][j][k-a[i][j]] + dp[i+1][j-1][k-a[i][j]];

        LL ans = 0;
        for(int i = 1; i <= n; ++i)  ans += dp[1][i][m];
        printf("%lld
", ans);
        if(ans)   print(1, -1, m);
        printf("
");
    }
    return 0;
}