BZOJ 1001 狼抓兔子 (最小割转化成最短路)
2017-10-02 23:30
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题意:中文题。
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:
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; }
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