Portal: http://www.lydsy.com/JudgeOnline/problem.php?id=1001
By the way recommend a PPT, there is a description of the floor plan: analysis of the maximum minimum theorem in the application of the information contest.
Here directly to the minimum cut must be T, so should be the original image as a floor plan, PPT said that the plane map corresponding to each of the dual map of the original one cut, which some do not understand, but do not affect this problem. Imagine, in the outermost of the infinite plane, from the upper left corner to the lower right corner with an additional edge, so do more an additional surface, set this additional edge of the weight of-inf, then the minimum cut must contain this edge. By removing this edge, it becomes a problem of finding the shortest path.
#include <cstdio> #include <cstring>const int maxn = 1005, maxnd = MAXN * MAXN << 1, maxe = MAXN * MAXN * 3;int N, M, S, T, special = 2147483647, T1, T2, T3;int head[maxnd], to[maxe << 1], next[maxe << 1], W[maxe &L t;< 1], Lb;char ch;bool inq[maxnd];int que[maxnd], H, head_, tail, d[maxnd];inline void ist (int aa, int ss, int ww) {to [lb] = ss;next[lb] = Head[aa];head[aa] = lb;w[lb] = ww;++lb;} inline void readint (int & RT) {while ((ch = getchar ()) <), RT = Ch-48;while ((ch = getchar ()) > +) {rt = RT * ten + ch-48;}} int main (void) {//freopen ("In.txt", "R", stdin), memset (head,-1, sizeof head), memset (Next,-1, sizeof next), Readint (n); r Eadint (m); if (n = = 1 | | m = = 1) {while (scanf ("%d", &t1)! = EOF) {special = Special < T1? Special:t1;} printf ("%d\n", special); return 0;} S = (n-1) * (m-1) * 2 + 1; T = S + 1;for (int i = 1; I <= n; ++i) {for (int j = 1; j < m; ++j) {t2 = (i-1) * (m-1) * 2 + J * 2;t1 = T2-(m -1) * 2-1;T1 = T1 > 0? T1:t;t2 = T2 < S? T2:S;SCANF ("%d", &t3), ist (t1, T2, T3), ist (T2, T1, T3);}} for (int i = 1; i < n; ++i) {t2 = (i-1) * (m-1) * 2 + 1;scanf ("%d", &t3), ist (S, t2, T3), ist (T2, S, T3), and for (int j = 2; J < M; ++J) {t1 = (i-1) * (m-1) * 2 + (j-1) * 2;t2 = t1 + 1;scanf ("%d", &t3), ist (t1, T2, T3), ist (T2, T1, T3); T1 = i * (m-1) * 2;SCANF ("%d", &t3), ist (t1, T, T3), ist (t, T1, T3);} for (int i = 1; i < n; ++i) {for (int j = 1; j < m; ++j) {t2 = (i-1) * (m-1) * 2 + J * 2;t1 = t2-1;scanf ("%d", &T3), ist (t1, T2, T3), ist (T2, T1, T3);}} memset (d, 0x3c, sizeof D); que[tail++] = s;inq[s] = True;d[s] = 0;while (head_! = tail) {h = que[head_++];if (Head_ = = T) { Head_ = 0;} INQ[H] = false;for (int j = head[h]; J! =-1; j = Next[j]) {if (D[to[j]] > D[h] + w[j]) {D[to[j]] = D[h] + w[j];if (!in Q[TO[J]]) {Inq[to[j]] = true;que[tail++] = to[j];if (tail = = T) {tail = 0;}}}} printf ("%d\n", D[t]); return 0;}
_bzoj1001 [BeiJing2006] wolf Scratch Rabbit "Floor plan"