洛谷P5319 奥术神杖

题意：给你若干个串和一个填了一部分的串。补完这个串使得 (每个串的匹配次数 * 权值) ^ (1 / 所有串匹配次数) 最大。

解：把这个东西随便取一个对数，就变成了分数规划。

二分。然后在AC自动机上DP判定。

#include <bits/stdc++.h>

const int N = 1510;
const double INF = 1e10, eps = 1e-16;

int tr[N][26], tot(1), cnt[N], g[N][N], gt[N][N], tans, fail[N], n;
double ed[N], f[N][N];
char str[N], ss[N];

inline void getfail() {

    std::queue<int> Q;
    Q.push(1);
    fail[1] = 1;
    while(Q.size()) {
        int x = Q.front();
        if(x != 1) {
            ed[x] += ed[fail[x]];
            cnt[x] += cnt[fail[x]];
        }
        Q.pop();
        for(int f = 0; f < 10; f++) {
            if(tr[x][f]) {
                int y = tr[x][f];
                if(x == 1) fail[y] = 1;
                else fail[y] = tr[fail[x]][f];
                Q.push(y);
            }
            else {
                if(x == 1) tr[x][f] = 1;
                else tr[x][f] = tr[fail[x]][f];
            }
        }
    }
    return;
}

inline void insert(double v) {
    int len = strlen(ss), p = 1;
    for(int i = 0; i < len; i++) {
        int f = ss[i] - '0';
        if(!tr[p][f]) {
            tr[p][f] = ++tot;
        }
        p = tr[p][f];
    }
    ed[p] += v;
    cnt[p]++;
    return;
}

inline double check(double D) {

    /// f[i][j] means len i in node j , max value

    for(int i = 0; i <= n; i++) {
        for(int j = 1; j <= tot; j++) {
            f[i][j] = -INF;
        }
    }

    double ans = -INF;
    f[0][1] = 0;
    for(int i = 0; i < n; i++) {
        for(int j = 1; j <= tot; j++) {
            //if(sgn(f[i][j] + INF) == 0) continue;
            if(f[i][j] == -INF) continue;
            //printf("%lf ", f[i][j]);
            if(str[i + 1] != '.') {
                int ff = str[i + 1] - '0';
                int y = tr[j][ff];
                if(f[i + 1][y] < f[i][j] + ed[y] - D * cnt[y]) {
                    f[i + 1][y] = f[i][j] + ed[y] - D * cnt[y];
                    g[i + 1][y] = j;
                    gt[i + 1][y] = ff;
                }
            }
            else {
                for(int ff = 0; ff < 10; ff++) {
                    int y = tr[j][ff];
                    //f[i + 1][y] = std::max(f[i + 1][y], f[i][j] + ed[y] - D * cnt[y]);
                    if(f[i + 1][y] < f[i][j] + ed[y] - D * cnt[y]) {
                        //printf("ff = %d  y = %d  
", ff, y);
                        f[i + 1][y] = f[i][j] + ed[y] - D * cnt[y];
                        g[i + 1][y] = j;
                        gt[i + 1][y] = ff;
                    }
                }
            }
        }
        //puts("");
    }

    for(int j = 1; j <= tot; j++) {
        if(ans < f[n][j] + eps) {
            ans = f[n][j];
            tans = j;
        }
    }

    return ans;
}

int main() {

    //freopen("my.out", "w", stdout);

    int m;
    double l = 0, r = 0;
    scanf("%d%d", &n, &m);
    scanf("%s", str + 1);
    double x;
    for(int i = 1; i <= m; i++) {
        scanf("%s%lf", ss, &x);
        //printf("i = %d 
", i);
        x = log(x);
        //printf(" h  21 1 1 
");
        insert(x);
        //printf(" fdsgfdsfsdfsh  21 1 1  
");
        r += x;
        int len = strlen(ss);
        memset(ss, 0, len * sizeof(char));
    }
    //printf("gfsiofdsfsdgfs
");

    getfail();

    for(int T = 1; T <= 30; T++) {
        double mid = (l + r) / 2;
        //printf("mid %.10f 
", mid);
        double t = check(mid);
        //printf("l = %lf r = %lf mid = %lf 
", l, r, mid);

        if(t > 0) {
            l = mid;
        }
        else {
            r = mid;
        }
    }

    //printf("r = %.10f 
", r);
    /// output ways
/*
6 5
2....2
252 62
5225 18
25 7
552 12
2122 18

*/

    check(l);
    for(int i = n; i >= 1; i--) {
        //printf("ans : %d 
", gt[i][tans]);
        //printf("node : j = %d 
", tans);
        str[i] = gt[i][tans];
        tans = g[i][tans];

    }

    for(int i = 1; i <= n; i++) {
        putchar(str[i] + '0');
    }
    /*for(int i = 1; i <= n; i++) {
        printf("%d 
", str[i]);
    }*/

    /*puts("");
    for(int i = 1; i <= tot; i++) {
        printf("i = %d fail = %d | ", i, fail[i]);
        for(int j = 0; j < 10; j++) {
            printf("%d ", tr[i][j]);
        }
        puts("");
    }*/

    return 0;
}

输出答案前如果不加那一句check(l)，会WA第二个点。