BZOJ2595 [Wc2008]游览计划 【状压dp + 最短路】

题目链接

BZOJ2595

题解

著名的斯坦纳树问题

\(f[i][j][s]\)表示点\((i,j)\)与景点联通状况为\(s\)的最小志愿者数
\(val[i][j]\)\((i,j)\)需要的志愿者数

有两种转移
一种是自己转移

\[f[i][j][s] = min\{f[i][j][e] + f[i][j][\complement_s e] - val[i][j]\} \]

一种是由周围转移过来

\[f[i][j][s] = min\{f[i][j][s] + f[x][y][s] + dis\} \]

第一种\(O(3^{K})\)枚举子集,第二种就是最短路

纪念一下BZOJ500题,截个图,,数字挺整的

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 11,maxm = 1 << 10,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int n,m,K,maxv;
int f[maxn][maxn][maxm],val[maxn][maxn],id[maxn][maxn],vis[maxn][maxn];
int head,tail,X[4] = {0,0,-1,1},Y[4] = {-1,1,0,0};
int S[maxn][maxn];
cp q[10000];
struct tri{
	int x,y,s;
};
vector<tri> pre[maxn][maxn][maxm];
void spfa(int s){
	head = 0; tail = -1;
	REP(i,n) REP(j,m) q[++tail] = mp(i,j),vis[i][j] = true;
	cp u; int nx,ny;
	while (head <= tail){
		u = q[head++];
		vis[u.first][u.second] = false;
		for (int k = 0; k < 4; k++){
			nx = u.first + X[k];
			ny = u.second + Y[k];
			if (nx < 1 || ny < 1 || nx > n || ny > m) continue;
			if (f[nx][ny][s] > f[u.first][u.second][s] + val[nx][ny]){
				f[nx][ny][s] = f[u.first][u.second][s] + val[nx][ny];
				pre[nx][ny][s].clear();
				pre[nx][ny][s].push_back((tri){u.first,u.second,s});
				if (!vis[nx][ny]) q[++tail] = mp(nx,ny);
			}
		}
	}
}
void dfs(int x,int y,int s){
	S[x][y] = true;
	tri u;
	for (unsigned int i = 0; i < pre[x][y][s].size(); i++){
		u = pre[x][y][s][i];
		dfs(u.x,u.y,u.s);
	}
}
void work(){
	maxv = (1 << K) - 1;
	memset(f,0x3f3f3f3f,sizeof(f));
	REP(i,n) REP(j,m){
		f[i][j][0] = 0;
		if (!val[i][j]) f[i][j][1 << id[i][j] - 1] = 0;
	}
	for (int s = 0; s <= maxv; s++){
		REP(i,n) REP(j,m){
			for (int e = s; e; e = (e - 1) & s){
				if (f[i][j][s] > f[i][j][e] + f[i][j][s ^ e] - val[i][j]){
					f[i][j][s] = f[i][j][e] + f[i][j][s ^ e] - val[i][j];
					pre[i][j][s].clear();
					pre[i][j][s].push_back((tri){i,j,e});
					pre[i][j][s].push_back((tri){i,j,s ^ e});
				}
			}
		}
		spfa(s);
	}
	int ans = INF,x,y;
	REP(i,n) REP(j,m) if (ans > f[i][j][maxv]) ans = f[i][j][maxv],x = i,y = j;
	printf("%d\n",ans);
	dfs(x,y,maxv);
	REP(i,n){
		REP(j,m){
			if (id[i][j]) putchar('x');
			else if (S[i][j]) putchar('o');
			else putchar('_');
		}
		puts("");
	}
}
int main(){
	n = read(); m = read();
	REP(i,n) REP(j,m){
		val[i][j] = read();
		if (!val[i][j]) id[i][j] = ++K;
	}
	work();
	return 0;
}

posted @ 2018-06-17 10:36  Mychael  阅读(109)  评论(0编辑  收藏  举报