Codeforces 295C Greg and Friends BFS

Greg and Friends

BFS的过程中维护一下方案数。 我个人感觉不是很好想, 但是写出来之后怎么感觉这题这么SB啊啊。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 50 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int n, k, cnt[2];
int dp[N][N][2];
LL comb[N][N];
LL ans[N][N][2];

int main() {
    for(int i = 0; i < N; i++)
        for(int j = comb[i][0] = 1; j <= i; j++)
            comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % mod;
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) {
        int x; scanf("%d", &x);
        if(x == 50) cnt[0]++;
        else cnt[1]++;
    }
    memset(dp, -1, sizeof(dp));
    queue<pair<PII, int>> que;
    que.push(mk(mk(0, 0), 0));
    ans[0][0][0] = 1;
    dp[0][0][0] = 0;
    while(!que.empty()) {
        int x = que.front().fi.fi;
        int y = que.front().fi.se;
        int op = que.front().se;
        que.pop();
        if(op) {
            for(int i = 0; i <= min(k / 50, x); i++) {
                int up = min(y, (k - i * 50) / 100);
                for(int j = 0; j <= up; j++) {
                    if(!i && !j) continue;
                    if(dp[x - i][y - j][0] == -1 || dp[x - i][y - j][0] == dp[x][y][1] + 1) {
                        if(dp[x - i][y - j][0] == -1) que.push(mk(mk(x - i, y - j), op ^ 1));
                        dp[x - i][y - j][0] = dp[x][y][1] + 1;
                        ans[x - i][y - j][0] = (ans[x - i][y - j][0] + ans[x][y][1] * comb[x][i] % mod * comb[y][j] % mod) % mod;
                    }
                }
            }
        } else {
            for(int i = 0; i <= min(k / 50, cnt[0] - x); i++) {
                int up = min(cnt[1] - y, (k - i * 50) / 100);
                for(int j = 0; j <= up; j++) {
                    if(!i && !j) continue;
                    if(dp[x + i][y + j][1] == -1 || dp[x + i][y + j][1] == dp[x][y][0] + 1) {
                        if(dp[x + i][y + j][1] == -1) que.push(mk(mk(x + i, y + j), op ^ 1));
                        dp[x + i][y + j][1] = dp[x][y][0] + 1;
                        ans[x + i][y + j][1] = (ans[x + i][y + j][1] + ans[x][y][0] * comb[cnt[0]-x][i] % mod * comb[cnt[1]-y][j] % mod) % mod;
                    }
                }
            }
        }
    }
    if(dp[cnt[0]][cnt[1]][1] == -1) {
        puts("-1");
        puts("0");
    } else {
        printf("%d\n", dp[cnt[0]][cnt[1]][1]);
        printf("%lld\n", ans[cnt[0]][cnt[1]][1]);
    }
    return 0;
}

/*
*/

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转载自www.cnblogs.com/CJLHY/p/10446939.html