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。
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 #include <queue> 5 using namespace std; 6 7 typedef long long LL; 8 #define maxn 55 9 #define INF 0xffffffff 10 const LL mod = 1000000007; 11 12 struct Point{ 13 int f, t, s; 14 Point(){} 15 Point(int _f, int _t, int _s): 16 f(_f), t(_t), s(_s){} 17 }; 18 19 LL dp[maxn][maxn][2]; 20 LL C[maxn][maxn]; 21 int hash[maxn][maxn][2]; 22 23 int bfs(int nf, int nt, int cap){ 24 queue<Point> Q; 25 dp[nf][nt][0] = 1; 26 hash[nf][nt][0] = 0; 27 Q.push(Point(nf, nt, 0)); 28 while(!Q.empty()){ 29 Point c = Q.front();Q.pop(); 30 int cf = c.f, ct = c.t, cs = c.s; 31 for(int i = 0; i <= c.f; i ++){ 32 if(i*50 > cap) break; 33 for(int j = 0; j <= c.t; j ++){ 34 if(i*50 + j*100 > cap) continue; 35 if(i*50 + j*100 <= 0) continue; 36 37 int lf = c.f - i, lt = c.t - j; 38 if(hash[nf - lf][nt - lt][!cs]==INF){ 39 hash[nf - lf][nt - lt][!cs] = hash[cf][ct][cs] + 1; 40 LL tmp = C[cf][i]*C[ct][j]%mod; 41 dp[nf - lf][nt - lt][!cs] += (tmp*dp[cf][ct][cs]%mod); 42 dp[nf - lf][nt - lt][!cs]%=mod; 43 Q.push(Point(nf-lf, nt-lt, !cs)); 44 } 45 else if(hash[nf - lf][nt - lt][!cs]==hash[cf][ct][cs] + 1){ 46 LL tmp = C[cf][i]*C[ct][j]%mod; 47 dp[nf - lf][nt - lt][!cs] += (tmp*dp[cf][ct][cs]%mod); 48 dp[nf - lf][nt - lt][!cs]%=mod; 49 } 50 } 51 } 52 } 53 return hash[nf][nt][1]; 54 } 55 56 void init(){ 57 C[0][0] = 1; 58 for(int i = 1; i <= 50; i ++){ 59 C[i][0] = 1; 60 for(int j = 1; j <= i; j ++){ 61 C[i][j] = (C[i-1][j-1] + C[i-1][j])%mod; 62 } 63 } 64 } 65 66 int main() 67 { 68 init(); 69 //freopen("test.in", "r", stdin); 70 for(int n, k, nf, nt; scanf("%d%d", &n, &k)!=EOF; ){ 71 memset(dp, 0, sizeof(dp)); 72 memset(hash, 0xff, sizeof(hash)); 73 nf = nt = 0; 74 for(int i = 0, x; i < n; i ++){ 75 scanf("%d", &x); 76 if(x==50) nf ++; 77 else nt ++; 78 } 79 bfs(nf, nt, k); 80 printf("%d %I64d ", hash[nf][nt][1], dp[nf][nt][1]); 81 } 82 return 0; 83 }