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。

  1 #include <bits/stdc++.h>
  2 
  3 const int N = 1510;
  4 const double INF = 1e10, eps = 1e-16;
  5 
  6 int tr[N][26], tot(1), cnt[N], g[N][N], gt[N][N], tans, fail[N], n;
  7 double ed[N], f[N][N];
  8 char str[N], ss[N];
  9 
 10 inline void getfail() {
 11 
 12     std::queue<int> Q;
 13     Q.push(1);
 14     fail[1] = 1;
 15     while(Q.size()) {
 16         int x = Q.front();
 17         if(x != 1) {
 18             ed[x] += ed[fail[x]];
 19             cnt[x] += cnt[fail[x]];
 20         }
 21         Q.pop();
 22         for(int f = 0; f < 10; f++) {
 23             if(tr[x][f]) {
 24                 int y = tr[x][f];
 25                 if(x == 1) fail[y] = 1;
 26                 else fail[y] = tr[fail[x]][f];
 27                 Q.push(y);
 28             }
 29             else {
 30                 if(x == 1) tr[x][f] = 1;
 31                 else tr[x][f] = tr[fail[x]][f];
 32             }
 33         }
 34     }
 35     return;
 36 }
 37 
 38 inline void insert(double v) {
 39     int len = strlen(ss), p = 1;
 40     for(int i = 0; i < len; i++) {
 41         int f = ss[i] - '0';
 42         if(!tr[p][f]) {
 43             tr[p][f] = ++tot;
 44         }
 45         p = tr[p][f];
 46     }
 47     ed[p] += v;
 48     cnt[p]++;
 49     return;
 50 }
 51 
 52 inline double check(double D) {
 53 
 54     /// f[i][j] means len i in node j , max value
 55 
 56     for(int i = 0; i <= n; i++) {
 57         for(int j = 1; j <= tot; j++) {
 58             f[i][j] = -INF;
 59         }
 60     }
 61 
 62     double ans = -INF;
 63     f[0][1] = 0;
 64     for(int i = 0; i < n; i++) {
 65         for(int j = 1; j <= tot; j++) {
 66             //if(sgn(f[i][j] + INF) == 0) continue;
 67             if(f[i][j] == -INF) continue;
 68             //printf("%lf ", f[i][j]);
 69             if(str[i + 1] != '.') {
 70                 int ff = str[i + 1] - '0';
 71                 int y = tr[j][ff];
 72                 if(f[i + 1][y] < f[i][j] + ed[y] - D * cnt[y]) {
 73                     f[i + 1][y] = f[i][j] + ed[y] - D * cnt[y];
 74                     g[i + 1][y] = j;
 75                     gt[i + 1][y] = ff;
 76                 }
 77             }
 78             else {
 79                 for(int ff = 0; ff < 10; ff++) {
 80                     int y = tr[j][ff];
 81                     //f[i + 1][y] = std::max(f[i + 1][y], f[i][j] + ed[y] - D * cnt[y]);
 82                     if(f[i + 1][y] < f[i][j] + ed[y] - D * cnt[y]) {
 83                         //printf("ff = %d  y = %d  \n", ff, y);
 84                         f[i + 1][y] = f[i][j] + ed[y] - D * cnt[y];
 85                         g[i + 1][y] = j;
 86                         gt[i + 1][y] = ff;
 87                     }
 88                 }
 89             }
 90         }
 91         //puts("");
 92     }
 93 
 94     for(int j = 1; j <= tot; j++) {
 95         if(ans < f[n][j] + eps) {
 96             ans = f[n][j];
 97             tans = j;
 98         }
 99     }
100 
101     return ans;
102 }
103 
104 int main() {
105 
106     //freopen("my.out", "w", stdout);
107 
108     int m;
109     double l = 0, r = 0;
110     scanf("%d%d", &n, &m);
111     scanf("%s", str + 1);
112     double x;
113     for(int i = 1; i <= m; i++) {
114         scanf("%s%lf", ss, &x);
115         //printf("i = %d \n", i);
116         x = log(x);
117         //printf(" h  21 1 1 \n");
118         insert(x);
119         //printf(" fdsgfdsfsdfsh  21 1 1  \n");
120         r += x;
121         int len = strlen(ss);
122         memset(ss, 0, len * sizeof(char));
123     }
124     //printf("gfsiofdsfsdgfs\n");
125 
126     getfail();
127 
128     for(int T = 1; T <= 30; T++) {
129         double mid = (l + r) / 2;
130         //printf("mid %.10f \n", mid);
131         double t = check(mid);
132         //printf("l = %lf r = %lf mid = %lf \n", l, r, mid);
133 
134         if(t > 0) {
135             l = mid;
136         }
137         else {
138             r = mid;
139         }
140     }
141 
142     //printf("r = %.10f \n", r);
143     /// output ways
144 /*
145 6 5
146 2....2
147 252 62
148 5225 18
149 25 7
150 552 12
151 2122 18
152 
153 */
154 
155     check(l);
156     for(int i = n; i >= 1; i--) {
157         //printf("ans : %d \n", gt[i][tans]);
158         //printf("node : j = %d \n", tans);
159         str[i] = gt[i][tans];
160         tans = g[i][tans];
161 
162     }
163 
164     for(int i = 1; i <= n; i++) {
165         putchar(str[i] + '0');
166     }
167     /*for(int i = 1; i <= n; i++) {
168         printf("%d \n", str[i]);
169     }*/
170 
171     /*puts("");
172     for(int i = 1; i <= tot; i++) {
173         printf("i = %d fail = %d | ", i, fail[i]);
174         for(int j = 0; j < 10; j++) {
175             printf("%d ", tr[i][j]);
176         }
177         puts("");
178     }*/
179 
180     return 0;
181 }