1001: [BeiJing2006]狼抓兔子
Time Limit: 15 Sec Memory Limit: 162 MB
Submit: 27715 Solved: 7134
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Description
现在小朋友们最喜欢的"喜羊羊与灰太狼",话说灰太狼抓羊不到，但抓兔子还是比较在行的，
而且现在的兔子还比较笨，它们只有两个窝，现在你做为狼王，面对下面这样一个网格的地形：
左上角点为(1,1),右下角点为(N,M)(上图中N=4,M=5).有以下三种类型的道路
1:(x,y)<==>(x+1,y)
2:(x,y)<==>(x,y+1)
3:(x,y)<==>(x+1,y+1)
道路上的权值表示这条路上最多能够通过的兔子数，道路是无向的. 左上角和右下角为兔子的两个窝，
开始时所有的兔子都聚集在左上角(1,1)的窝里，现在它们要跑到右下解(N,M)的窝中去，狼王开始伏击
这些兔子.当然为了保险起见，如果一条道路上最多通过的兔子数为K，狼王需要安排同样数量的K只狼，
才能完全*这条道路，你需要帮助狼王安排一个伏击方案，使得在将兔子一网打尽的前提下，参与的
狼的数量要最小。因为狼还要去找喜羊羊麻烦.
Input
第一行为N,M.表示网格的大小，N,M均小于等于1000.
接下来分三部分
第一部分共N行，每行M-1个数，表示横向道路的权值.
第二部分共N-1行，每行M个数，表示纵向道路的权值.
第三部分共N-1行，每行M-1个数，表示斜向道路的权值.
输入文件保证不超过10M
Output
输出一个整数，表示参与伏击的狼的最小数量.
Sample Input
3 4
5 6 4
4 3 1
7 5 3
5 6 7 8
8 7 6 5
5 5 5
6 6 6
5 6 4
4 3 1
7 5 3
5 6 7 8
8 7 6 5
5 5 5
6 6 6
Sample Output
14
HINT
2015.4.16新加数据一组，可能会卡掉从前可以过的程序。
Source
析：很容易看出是裸板的最小割，然后可能会超时，边实在是太多了，有一种特殊的方法，可以把平面图转成最短路来求，也就是利用对偶图，把原图的而看成新图的点，原图的边与两个面相连的，加一条边，然后再多加一个起点和终点。跑一次最短路即可。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
//#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e6 + 10;
const int maxm = 100 + 10;
const ULL mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
int to, dist, next;
}; struct HeapNode{
int d, u;
bool operator < (const HeapNode &p) const{
return d > p.d;
}
};
Edge edge[maxn*3];
int head[maxn], cnt;
int d[maxn];
bool done[maxn]; inline void addEdge(int u, int v, int dist){
edge[cnt].to = v;
edge[cnt].dist = dist;
edge[cnt].next = head[u];
head[u] = cnt++;
} void dijkstra(int s){
priority_queue<HeapNode> pq;
ms(d, INF); d[s] = 0;
ms(done, 0);
pq.push((HeapNode){0, s});
while(!pq.empty()){
HeapNode x = pq.top(); pq.pop();
int u = x.u;
if(done[u]) continue;
done[u] = 1;
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to, dist = edge[i].dist;
if(d[v] > d[u] + dist){
d[v] = d[u] + dist;
pq.push((HeapNode){d[v], v});
}
}
}
} int main(){
scanf("%d %d", &n, &m);
if(n == 1 || m == 1){
n = max(n, m);
int ans = INF;
for(int i = 0; i < n; ++i){
int x;
scanf("%d", &x);
ans = min(ans, x);
}
printf("%d\n", ans);
return 0;
}
ms(head, -1); cnt = 0;
int s = 0, t = 2 * n * m + 1;
FOR(i, 0, n) for(int j = 1; j < (m-1<<1); j += 2){
int dist;
scanf("%d", &dist);
int from = i == 0 ? s : (i-1)*(m-1<<1)+j+1;
int to = i + 1 == n ? t : i*(m-1<<1)+j;
addEdge(from, to, dist);
addEdge(to, from, dist);
}
FOR(i, 0, n-1) for(int j = 1; j <= m; ++j){
int dist;
scanf("%d", &dist);
int from = j == 1 ? t : i*(m-1<<1)+(j<<1)-3;
int to = j == m ? s : i*(m-1<<1)+(j<<1);
addEdge(from, to, dist);
addEdge(to, from, dist);
}
FOR(i, 0, n-1) for(int j = 1; j < m; ++j){
int dist;
scanf("%d", &dist);
int from = i*(m-1<<1)+(j<<1);
int to = from - 1;
addEdge(from, to, dist);
addEdge(to, from, dist);
}
dijkstra(s);
printf("%d\n", d[t]);
return 0;
}