题目链接

BZOJ2007

题解

这是裸题啊，，要是考试真的遇到就好了

明显是最小割，而且是有来回两个方向

那么原图所有向右的边转为对偶图向下的边

向左的边转为向上

向下转为向左

向上转为向右

然后跑一遍最短路即可

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<vector>

#include<queue>

#include<cmath>

#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<LL,int>(a,b)

#define cls(s) memset(s,0,sizeof(s))

#define cp pair<LL,int>

#define LL long long int

#define id(x,y) (n * (x - 1) + y)

using namespace std;

const int maxn = 300005,maxm = 10000005;

const LL INF = 1000000000000000ll;

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 h[maxn],ne;

struct EDGE{int to,nxt,w;}ed[maxm];

inline void build(int u,int v,int w){

	ed[++ne] = (EDGE){v,h[u],w}; h[u] = ne;

}

int n,N,S,T,vis[maxn];

LL d[maxn];

priority_queue<cp,vector<cp>,greater<cp> > q;

void dijkstra(){

	for (int i = 1; i <= N; i++) d[i] = INF; d[S] = 0;

	q.push(mp(d[S],S));

	int u;

	while (!q.empty()){

		u = q.top().second; q.pop();

		if (vis[u]) continue;

		vis[u] = true;

		Redge(u) if (!vis[to = ed[k].to] && d[to] > d[u] + ed[k].w){

			d[to] = d[u] + ed[k].w;

			q.push(mp(d[to],to));

		}

	}

}

void readin(){

	for (int i = 1; i <= n; i++) build(S,id(1,i),read());

	for (int i = 1; i < n; i++){

		for (int j = 1; j <= n; j++)

			build(id(i,j),id(i + 1,j),read());

	}

	for (int i = 1; i <= n; i++) build(id(n,i),T,read());

	for (int i = 1; i <= n; i++){

		build(id(i,1),T,read());

		for (int j = 1; j < n; j++)

			build(id(i,j + 1),id(i,j),read());

		build(S,id(i,n),read());

	}

	for (int i = 1; i <= n; i++) build(id(1,i),S,read());

	for (int i = 1; i < n; i++){

		for (int j = 1; j <= n; j++)

			build(id(i + 1,j),id(i,j),read());

	}

	for (int i = 1; i <= n; i++) build(T,id(n,i),read());

	for (int i = 1; i <= n; i++){

		build(T,id(i,1),read());

		for (int j = 1; j < n; j++)

			build(id(i,j),id(i,j + 1),read());

		build(id(i,n),S,read());

	}

}

int main(){

	n = read(); N = n * n + 2; S = N - 1; T = N;

	readin();

	dijkstra();

	printf("%lld\n",d[T]);

	return 0;

}

巴特西

BZOJ2007 [Noi2010]海拔【平面图最小割转对偶图最短路】

题目链接

题解

最新文章

热门文章

巴特西

BZOJ2007 [Noi2010]海拔 【平面图最小割转对偶图最短路】

题目链接

题解

最新文章

热门文章

BZOJ2007 [Noi2010]海拔【平面图最小割转对偶图最短路】