洛谷T31039 九尾狐吃棉花糖

小伙伴出的题。

一眼看出是状压DP裸题。回忆poj2288 islands and bridges，然后就很好写了。

啪啪啪打了个状压DP出来(晚上寝室写的，其实是记忆化搜索)，发现sum总是INF

然后发现：printf函数调用的貌似不是运行solve之后的，而是还未运行solve时的值。

于是分开写就A了。

出题者跑了一秒多，貌似用的二分...反正不怎么看得懂。

果然DP大法吼哇！

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 typedef long long LL;
 using std::max;
 using std::min;
 const int N = ;
 LL INF;
 int n;

 LL f[ << N][N], sum[ << N][N], G[N][N];

 void cal(int sta) {
     int t[N];
     for(int i = ; i < n; i++) {
         t[i] = (sta >> i) & ;
     }
     for(int i = n - ; i >= ; i--) {
         printf("%d", t[i]);
     }
     return;
 }

 LL solve(int i, int j) {

     if(f[i][j] != INF) {
         //printf("solve: ");
         //cal(i);
         //printf(" %d ans = %lld sum = %lld \n", j, f[i][j], sum[i][j]);
         return f[i][j];
     }

     int ct = ;
     for(int ii = ; ii < n; ii++) {
         ct += (i >> ii) & ;
     }
     if(ct < ) {
         f[i][j] = INF - ;
         return INF - ;
     }

     LL ans = INF - , len = INF - ;
     int sta = ((~( << j)) & i);
     for(int ii = ; ii < n; ii++) {
         if(!(G[ii][j] && ((i >> ii) & ))) {
             continue;
         }
         LL t_ans = max(solve(sta, ii), G[ii][j]);
         LL t_len = sum[sta][ii] + G[ii][j];
         if(t_ans < ans) {
             ans = t_ans;
             len = t_len;
         }
         else if(ans == t_ans) {
             len = min(len, t_len);
         }
     }
     f[i][j] = ans;
     sum[i][j] = len;
     //printf("solve: ");
     //cal(i);
     //printf(" %d ans = %lld sum = %lld \n", j, ans, len);
     return ans;
 }

 int main() {
     memset(f, 0x7f, sizeof(f));
     memset(sum, 0x7f, sizeof(sum));
     INF = sum[][];

     int m, x, y;
     LL R, z;
     scanf("%d%d%lld", &n, &m, &R);
     for(int i = ; i <= m; i++) {
         scanf("%d%d%lld", &x, &y, &z);
         x--; y--;
         if(x == y || z > R) continue;
         if(G[x][y]) z = min(z, G[x][y]);
         G[x][y] = G[y][x] = z;
     }
     for(int i = ; i < n; i++) {
         if(G[][i]) {
             int sta =  | ( << i);
             f[sta][i] = G[][i];
             sum[sta][i] = G[][i];
         }
     }
     int ed = ( << n) - ;
     LL ans = solve(ed, (n - ));
     LL len = sum[ed][n - ];
     if(ans >= INF - ) {
         printf("wuwuwu~");
     }
     else {
         printf("%lld %lld", ans, len);
     }
     return ;
 }

AC代码