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 = ;
 const double INF = 1e10, eps = 1e-;

 int tr[N][], tot(), 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();
     fail[] = ;
     while(Q.size()) {
         int x = Q.front();
         if(x != ) {
             ed[x] += ed[fail[x]];
             cnt[x] += cnt[fail[x]];
         }
         Q.pop();
         for(int f = ; f < ; f++) {
             if(tr[x][f]) {
                 int y = tr[x][f];
                 if(x == ) fail[y] = ;
                 else fail[y] = tr[fail[x]][f];
                 Q.push(y);
             }
             else {
                 if(x == ) tr[x][f] = ;
                 else tr[x][f] = tr[fail[x]][f];
             }
         }
     }
     return;
 }

 inline void insert(double v) {
     int len = strlen(ss), p = ;
     for(int i = ; i < len; i++) {
         int f = ss[i] - '';
         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 = ; i <= n; i++) {
         for(int j = ; j <= tot; j++) {
             f[i][j] = -INF;
         }
     }

     double ans = -INF;
     f[][] = ;
     for(int i = ; i < n; i++) {
         for(int j = ; 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 + ] != '.') {
                 int ff = str[i + ] - '';
                 int y = tr[j][ff];
                 if(f[i + ][y] < f[i][j] + ed[y] - D * cnt[y]) {
                     f[i + ][y] = f[i][j] + ed[y] - D * cnt[y];
                     g[i + ][y] = j;
                     gt[i + ][y] = ff;
                 }
             }
             else {
                 for(int ff = ; ff < ; 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 + ][y] < f[i][j] + ed[y] - D * cnt[y]) {
                         //printf("ff = %d  y = %d  \n", ff, y);
                         f[i + ][y] = f[i][j] + ed[y] - D * cnt[y];
                         g[i + ][y] = j;
                         gt[i + ][y] = ff;
                     }
                 }
             }
         }
         //puts("");
     }

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

     getfail();

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

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

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

 */

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

     }

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

AC代码

输出答案前如果不加那一句check(l)，会WA第二个点。