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[codeforces-543B]bfs求最短路

时间:2015-05-08 06:59:18      阅读:188      评论:0      收藏:0      [点我收藏+]

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题意:给一个边长为1的无向图,求删去最多的边使得从a到b距离<=f,从c到d距离<=g,a,b,c,d,f,g都是给定的,求最多删去的边数。

思路:反过来思考,用最少的边构造两条从a到b,从c到d的路径,使得它们满足题目中的条件。于是可以把这两条路径的相对位置分为两种情况,不相交和相交,对于不相交的情况答案就是两组距离之和,对于相交的情况,两条路径肯定共享一段连续路径,对于共享多段的情况可以把中间没共享的取较小值变成共享,得到的距离和不会更大,因此最优情况一定是一段连续的路径。于是可以枚举共享段的两个端点来更新答案,两点之间距离可以通过n次bfs求出。有个小地方需要注意,共享段对于两条路径来说是有方向的,所以需要正反更新。

技术分享
  1 #pragma comment(linker, "/STACK:10240000,10240000")
  2 
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <cstdlib>
  7 #include <cstring>
  8 #include <map>
  9 #include <queue>
 10 #include <deque>
 11 #include <cmath>
 12 #include <vector>
 13 #include <ctime>
 14 #include <cctype>
 15 #include <set>
 16 #include <bitset>
 17 #include <functional>
 18 #include <numeric>
 19 #include <stdexcept>
 20 #include <utility>
 21 
 22 using namespace std;
 23 
 24 #define mem0(a) memset(a, 0, sizeof(a))
 25 #define mem_1(a) memset(a, -1, sizeof(a))
 26 #define lson l, m, rt << 1
 27 #define rson m + 1, r, rt << 1 | 1
 28 #define define_m int m = (l + r) >> 1
 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 32 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 33 #define all(a) (a).begin(), (a).end()
 34 #define lowbit(x) ((x) & (-(x)))
 35 #define constructInt5(name, a, b, c, d, e) name(int a = 0, int b = 0, int c = 0, int d = 0, int e = 0): a(a), b(b), c(c), d(d), e(e) {}
 36 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 37 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 38 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 39 #define pchr(a) putchar(a)
 40 #define pstr(a) printf("%s", a)
 41 #define sstr(a) scanf("%s", a)
 42 #define sint(a) scanf("%d", &a)
 43 #define sint2(a, b) scanf("%d%d", &a, &b)
 44 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 45 #define pint(a) printf("%d\n", a)
 46 #define test_print1(a) cout << "var1 = " << a << endl
 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 49 #define mp(a, b) make_pair(a, b)
 50 #define pb(a) push_back(a)
 51 
 52 typedef long long LL;
 53 typedef pair<int, int> pii;
 54 typedef vector<int> vi;
 55 
 56 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
 57 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
 58 const int maxn = 3e3 + 7;
 59 const int md = 10007;
 60 const int inf = 1e9 + 7;
 61 const LL inf_L = 1e18 + 7;
 62 const double pi = acos(-1.0);
 63 const double eps = 1e-6;
 64 
 65 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
 66 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 67 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 68 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 69 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 70 int make_id(int x, int y, int n) { return x * n + y; }
 71 
 72 struct Graph {
 73     vector<vector<int> > G;
 74     void clear() { G.clear(); }
 75     void resize(int n) { G.resize(n + 2); }
 76     void add(int u, int v) { G[u].push_back(v); }
 77     vector<int> & operator [] (int u) { return G[u]; }
 78 };
 79 Graph G;
 80 
 81 int n, m, s1, s2, t1, t2, l1, l2, ans, dist[maxn][maxn];
 82 bool vis[maxn];
 83 
 84 void chk(int u, int v, int w) {
 85     int cost1 = dist[s1][u] + w + dist[v][t1], cost2 = dist[s2][u] + w + dist[v][t2], cost = cost1 + cost2 - w;
 86     if (cost1 <= l1 && cost2 <= l2) min_update(ans, cost);
 87     swap(u, v);
 88     cost2 = dist[s2][u] + w + dist[v][t2];
 89     cost = cost1 + cost2 - w;
 90     if (cost1 <= l1 && cost2 <= l2) min_update(ans, cost);
 91 }
 92 
 93 void bfs(int s) {
 94     queue<pii> q;
 95     q.push(make_pair(s, 0));
 96     mem0(vis);
 97     vis[s] = true;
 98     while (!q.empty()) {
 99         pii hnode = q.front(); q.pop();
100         int u = hnode.first, w = hnode.second;
101         dist[s][u] = w;
102         int sz = G[u].size();
103         rep_up0(i, sz) {
104             int v = G[u][i];
105             if (!vis[v]) {
106                 vis[v] = true;
107                 q.push(make_pair(v, w + 1));
108             }
109         }
110     }
111 }
112 
113 int main() {
114     //freopen("in.txt", "r", stdin);
115     cin >> n >> m;
116     memset(dist, 0x3f, sizeof(dist));
117     G.clear();
118     G.resize(n);
119     rep_up0(i, m) {
120         int u, v;
121         sint2(u, v);
122         G.add(u, v);
123         G.add(v, u);
124     }
125     sint3(s1, t1, l1);
126     sint3(s2, t2, l2);
127     rep_up1(i, n) {
128         bfs(i);
129     }
130     if (dist[s1][t1] > l1 || dist[s2][t2] > l2) {
131         puts("-1");
132         return 0;
133     }
134     ans = dist[s1][t1] + dist[s2][t2];
135     rep_up1(i, n) {
136         rep_up1(j, n) {
137             chk(i, j, dist[i][j]);
138         }
139     }
140     cout << m - ans << endl;
141     return 0;
142 }
View Code

 

[codeforces-543B]bfs求最短路

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原文地址:http://www.cnblogs.com/jklongint/p/4486629.html

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