Codeforces 295C Greg and Friends

BFS+DP.dp[i][j][0]表示有i个50kg，j个100kg的人在左岸，dp[i][j][1]表示有i个50kg,j个100kg的人在右岸。用BFS求最短路的时候记录到达该状态的可能情况。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;

typedef long long LL;
#define maxn 55
#define INF 0xffffffff
const LL mod = 1000000007;

struct Point{
    int f, t, s;
    Point(){}
    Point(int _f, int _t, int _s):
    f(_f), t(_t), s(_s){}
};

LL dp[maxn][maxn][2];
LL C[maxn][maxn];
int hash[maxn][maxn][2];

int bfs(int nf, int nt, int cap){
    queue<Point> Q;
    dp[nf][nt][0] = 1;
    hash[nf][nt][0] = 0;
    Q.push(Point(nf, nt, 0));
    while(!Q.empty()){
        Point c = Q.front();Q.pop();
        int cf = c.f, ct = c.t, cs = c.s;
        for(int i = 0; i <= c.f; i ++){
            if(i*50 > cap) break;
            for(int j = 0; j <= c.t; j ++){
                if(i*50 + j*100 > cap) continue;
                if(i*50 + j*100 <= 0) continue;

                int lf = c.f - i, lt = c.t - j;
                if(hash[nf - lf][nt - lt][!cs]==INF){
                    hash[nf - lf][nt - lt][!cs] = hash[cf][ct][cs] + 1;
                    LL tmp = C[cf][i]*C[ct][j]%mod;
                    dp[nf - lf][nt - lt][!cs] += (tmp*dp[cf][ct][cs]%mod);
                    dp[nf - lf][nt - lt][!cs]%=mod;
                    Q.push(Point(nf-lf, nt-lt, !cs));
                }
                else if(hash[nf - lf][nt - lt][!cs]==hash[cf][ct][cs] + 1){
                    LL tmp = C[cf][i]*C[ct][j]%mod;
                    dp[nf - lf][nt - lt][!cs] += (tmp*dp[cf][ct][cs]%mod);
                    dp[nf - lf][nt - lt][!cs]%=mod;
                }
            }
        }
    }
    return hash[nf][nt][1];
}

void init(){
    C[0][0] = 1;
    for(int i = 1; i <= 50; i ++){
        C[i][0] = 1;
        for(int j = 1; j <= i; j ++){
            C[i][j] = (C[i-1][j-1] + C[i-1][j])%mod;
        }
    }
}

int main()
{
    init();
    //freopen("test.in", "r", stdin);
    for(int n, k, nf, nt; scanf("%d%d", &n, &k)!=EOF; ){
        memset(dp, 0, sizeof(dp));
        memset(hash, 0xff, sizeof(hash));
        nf = nt = 0;
        for(int i = 0, x; i < n; i ++){
            scanf("%d", &x);
            if(x==50) nf ++;
            else nt ++;
        }
        bfs(nf, nt, k);
        printf("%d
%I64d
", hash[nf][nt][1], dp[nf][nt][1]);
    }
    return 0;
}