Bzoj 3232: Enclosure game 01 Score Planning

First two-point answer (Dinkelbach. Don't know if it's OK), and then it becomes possible to judge.

Here is a more ingenious way, is to set a direction for the enclosure, it may be counter-clockwise, then go upward is equivalent to the left side of the answer to add it, go down is equivalent to the left side of the answer to subtract, and then determine whether there is a negative power loop.

In addition, the network flow map, all blocks from the s edge capacity for value, and then add a circle to the t edge capacity of Benquan inf, and then any two adjacent blocks (including the outside lap) with the two-way edge capacity for the cost of the edge. Then run the minimum cut to subtract from the answer.

The AC code is as follows:

#include <iostream> #include <cstdio> #include <cstring> #define N #define M 100005 using namespace std

;
int m,n,cnt,h[m+5][2],s[n][n],mp[n][n][4],num[n][n]; Double d[n][n],len[n][n][4];
BOOL Bo[n][n];
	const int dx[4]={-1,1,0,0},dy[4]={0,0,-1,1};//0 on 1 2 left 3 right bool OK (double t) {int i,j,k,x,y;
				For (i=0, i<=m; i++) for (j=0; j<=n; j + +) for (k=0; k<4; k++) {x=i+dx[k]; y=j+dy[k];
					if (x>=0 && x<=m && y>=0 && y<=n) {len[i][j][k]=mp[i][j][k]*t; if (!k) len[i][j][k]-=s[i][j];
				else if (k==1) len[i][j][k]+=s[i+1][j];
	}} int head=0,tail=cnt,u=0,v=0;
		for (I=1; i<=cnt; i++) {h[i][0]=u; h[i][1]=v; if (v<n) v++;
	else{u++; v=0;} } memset (Bo,0,sizeof (bo));
	memset (num,0,sizeof (num));
		while (Head!=tail) {head=head%m+1; x=h[head][0]; y=h[head][1]; bo[x][y]=1;
			for (i=0; i<4; i++) {u=x+dx[i]; v=y+dy[i]; if (u>=0 && u<=m && v>=0 && v<=n) IF (D[x][y]+len[x][y][i]<d[u][v]) {d[u][v]=d[x][y]+len[x][y][i];
					num[u][v]=num[x][y]+1;
					if (num[u][v]>cnt) return 1;
						if (Bo[u][v]) {bo[u][v]=0; tail=tail%m+1; H[tail][0]=u;
					H[tail][1]=v;
}}}} return 0;
	} int main () {scanf ("%d%d", &m,&n); cnt= (m+1) * (n+1); int i,j,x;
		for (I=1; i<=m; i++) for (j=1; j<=n; J + +) {scanf ("%d", &x); s[i][j]=s[i][j-1]+x;
			} for (i=0; i<=m; i++) for (j=1; j<=n; J + +) {scanf ("%d", &x);
		Mp[i][j][2]=mp[i][j-1][3]=x;
			} for (I=1; i<=m; i++) for (j=0; j<=n; J + +) {scanf ("%d", &x);
		Mp[i][j][0]=mp[i-1][j][1]=x;
	} double L=0,r=m*n*100,mid;
		while (L+1e-6<r) {mid= (l+r)/2; if (OK (mid)) L=mid;
	else R=mid;
	} printf ("%.3f\n", (L+R)/2);
return 0;
 }

by Lych

2016.4.13