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 = ;

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

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

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

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

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