ZOJ-3103 cliff climbing Short Circuit

Meaning: arrive at the T point from several s points. The difference is that you need to distinguish between the left foot and the right foot.

Solution: Starting from the S given in the question, you can set foot on both the left and right feet, and enter the queue all before spfa. After doing this, we found that spfa is used to process multi-source points and no super source points need to be created.

The Code is as follows:

# Include <iostream> # include <cstring> # include <cstdlib> # include <cstdio> # include <queue> using namespace STD; const int INF = 0x3f3f3f; int n, m, dis [2] [65] [35]; char G [65] [35], vis [2] [65] [35]; struct point {int X, Y; point (int x _, int Y _) {x = X _, y = Y _;} Point () {}}; vector <point> S; bool judge (INT lx, int ly, int RX, int ry) {If (LX <1 | RX <1 | lx> N | RX> N | ly <1 | ry <1 | LY> M | Ry> m) return false; If (ly <ry & ABS (Lx-Rx) + ABS (ly-ry) <= 3) return true; return false ;} int ldir [9] [2] = {0,-1}, {0,-2}, {0,-3}, {1,-1 }, {1,-2}, {2,-1}, {-1,-1}, {-1,-2}, {-2,-1 }}; int rdir [9] [2] = {0, 1}, {0, 2}, {0, 3}, {1, 1}, {2, 1}, {-1 }, {-1, 2}, {-2, 1}; void spfa () {point V; memset (VIS, 0, sizeof (VIS); memset (DIS, 0x3f, sizeof (DIS); queue <point> q; For (INT I = 0; I! = S. size (); ++ I) {q. push (s [I]); DIS [0] [s [I]. x] [s [I]. y] = 0; DIS [1] [s [I]. x] [s [I]. y] = 0; vis [0] [s [I]. x] [s [I]. y] = 1; vis [1] [s [I]. x] [s [I]. y] = 1;} while (! Q. empty () {v = Q. front (); q. pop (); // first step left if (vis [1] [v. x] [v. y]) {// if the right foot is in the queue vis [1] [v. x] [v. y] = 0; For (int K = 0; k <9; ++ K) {int I = v. X + ldir [k] [0], j = v. Y + ldir [k] [1]; If (Judge (I, j, V. x, V. y) & G [I] [J]! = 'X') {int Ti = (G [I] [J] = 's' | G [I] [J] = 'T ')? 0: G [I] [J]-'0'; If (DIS [0] [I] [J]> dis [1] [v. x] [v. y] + Ti) {dis [0] [I] [J] = dis [1] [v. x] [v. y] + Ti; If (! Vis [0] [I] [J]) {q. push (point (I, j); vis [0] [I] [J] = 1 ;}}}}} // if you step on the right foot if (vis [0] [v. x] [v. y]) {vis [0] [v. x] [v. y] = 0; For (int K = 0; k <9; ++ K) {int I = v. X + rdir [k] [0], j = v. Y + rdir [k] [1]; If (Judge (v. x, V. y, I, j) & G [I] [J]! = 'X') {int Ti = (G [I] [J] = 's' | G [I] [J] = 'T ')? 0: G [I] [J]-'0'; If (DIS [1] [I] [J]> dis [0] [v. x] [v. y] + Ti) {dis [1] [I] [J] = dis [0] [v. x] [v. y] + Ti; If (! Vis [1] [I] [J]) {q. push (point (I, j); vis [1] [I] [J] = 1 ;}}}} int main () {While (CIN> m> N, N | M) {getchar (); S. clear (); For (INT I = 1; I <= N; ++ I) {for (Int J = 1; j <= m; ++ J) {scanf ("% C % * C", & G [I] [J]); If (G [I] [J] = 's') {S. push_back (point (I, j) ;}} spfa (); int min = inf; For (INT I = 1; I <= N; ++ I) {for (Int J = 1; j <= m; ++ J) {If (G [I] [J] = 'T') {min = min (Min, min (Di S [0] [I] [J], DIS [1] [I] [J]) ;}} if (Min! = Inf) {printf ("% d \ n", min);} else printf ("-1 \ n");} return 0 ;}