20190601

Author: Del-te

TangoTangoInsurrectionUVa10618.cpp

@@ -0,0 +1,101 @@
+#include<cstdio>
+#include<cstring>
+#include<cassert>
+
+const int UP = 0;
+const int LEFT = 1;
+const int RIGHT = 2;
+const int DOWN = 3;
+
+const int maxn = 70 + 5;
+
+// d[i][a][b][s] means the minimal future energy when you already tapped i notes
+// your left foot at a, right foot at b, last foot is s
+int d[maxn][4][4][3];
+
+// if the optimal strategy is to move foot f(0~2) to position t, action=f*4+t
+int action[maxn][4][4][3]; 
+
+char seq[maxn], pos[256], footch[] = ".LR";
+
+// energy needed to move a foot FOR THE SECOND TIME, from a to ta
+int energy(int a, int ta) {
+  if(a == ta) return 3;
+  if(a + ta == 3) return 7; // across
+  return 5; // adjacent
+}
+
+int energy(int i, int a, int b, int s, int f, int t, int& ta, int& tb) {
+  ta = a; tb = b;
+  if(f == 1) ta = t;
+  else if(f == 2) tb = t;
+
+  // check target arrows
+  if(ta == tb) return -1;
+  if(ta == RIGHT && tb == LEFT) return -1;
+  if(a == RIGHT && tb != b) return -1; // you can't move you right foot before your left foot comes back
+  if(b == LEFT  && ta != a) return -1;
+
+  // compute energy
+  int e;
+  if(f == 0) e = 0; // no move
+  else if(f != s) e = 1; // alternative foot, low energy
+  else {
+    if(f == 1) e = energy(a, ta);
+    else e = energy(b, tb);
+  }
+  return e;
+}
+
+// update state (i,a,b,s). foot f is moved to t
+void update(int i, int a, int b, int s, int f, int t) {
+  int ta, tb;
+  int e = energy(i, a, b, s, f, t, ta, tb);
+  if(e < 0) return; // invalid
+
+  int cost = d[i+1][ta][tb][f] + e;
+  int& ans = d[i][a][b][s];
+  if(cost < ans) {
+    ans = cost;
+    action[i][a][b][s] = f * 4 + t;
+  }
+}
+
+int main() {
+  pos['U'] = 0; pos['L'] = 1; pos['R'] = 2; pos['D'] = 3;
+
+  while(scanf("%s", seq) == 1) {
+    if(seq[0] == '#') break;
+    int n = strlen(seq);
+    memset(d, 0, sizeof(d));
+    for(int i = n-1; i >= 0; i--)
+      for(int a = 0; a < 4; a++)
+        for(int b = 0; b < 4; b++) if(a != b)
+          for(int s = 0; s < 3; s++) {
+            d[i][a][b][s] = 10*n;
+            if(seq[i] == '.') {
+              update(i, a, b, s, 0, 0); // no move
+              for(int t = 0; t < 4; t++) {
+                update(i, a, b, s, 1, t); // move left foot
+                update(i, a, b, s, 2, t); // move right foot
+              }
+            } else {
+              update(i, a, b, s, 1, pos[seq[i]]); // move left foot
+              update(i, a, b, s, 2, pos[seq[i]]); // move right foot
+            }
+          }
+
+    // print solution
+    int a = LEFT, b = RIGHT, s = 0; // d[0][1][2][0] is out answer
+    for(int i = 0; i < n; i++) {
+      int f = action[i][a][b][s] / 4;
+      int t = action[i][a][b][s] % 4;
+      printf("%c", footch[f]);
+      s = f;
+      if(f == 1) a = t;
+      else if(f == 2) b = t;
+    }
+    printf("\n");
+  }
+  return 0;
+}