HDU 5294 Tricks Device (最大流+最短路)
Posted llguanli
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题意:n个点,m条边。而且一个人从1走到n仅仅会走1到n的最短路径。问至少破坏几条边使原图的最短路不存在。最多破坏几条边使原图的最短路劲仍存在
思路:
1.跑一遍最短路。记录下全部最短路径,将这些最短路径的边以(0,1)(流量,容量)加到网络流中,跑一遍最大流
2.记录下的全部最短路径,再加到新的最短路的图中,边权为1,跑一遍最短路。m-最短路 就是答案
注意:有重边的情况
比方:
2 3
1 2 1
1 2 1
1 2 1
ans: 3 2
AC代码:
#include <stdio.h> #include <string.h> #include <algorithm> #include <queue> #include <set> using namespace std; const int MAXN=2010; const int MAXM=60010; #define typec int const typec INF=0x3f3f3f3f;//防止后面溢出,这个不能太大 struct Edge { int to,next,cap,flow; } edge[MAXM*4]; //注意是MAXM int tol; int head[MAXN]; int gap[MAXN],dep[MAXN],cur[MAXN]; void add(int u,int v,int w,int rw = 0) { edge[tol].to = v; edge[tol].cap = w; edge[tol].flow = 0; edge[tol].next = head[u]; head[u] = tol++; edge[tol].to = u; edge[tol].cap = rw; edge[tol].flow = 0; edge[tol].next = head[v]; head[v] = tol++; } int Q[MAXN]; void BFS(int start,int end) { memset(dep,-1,sizeof(dep)); memset(gap,0,sizeof(gap)); gap[0] = 1; int front = 0, rear = 0; dep[end] = 0; Q[rear++] = end; while(front != rear) { int u = Q[front++]; for(int i = head[u]; i != -1; i = edge[i].next) { int v = edge[i].to; if(dep[v] != -1)continue; Q[rear++] = v; dep[v] = dep[u] + 1; gap[dep[v]]++; } } } int S[MAXN]; int sap(int start,int end,int N) { BFS(start,end); memcpy(cur,head,sizeof(head)); int top = 0; int u = start; int ans = 0; while(dep[start] < N) { if(u == end) { int Min = INF; int inser; for(int i = 0; i < top; i++) if(Min > edge[S[i]].cap - edge[S[i]].flow) { Min = edge[S[i]].cap - edge[S[i]].flow; inser = i; } for(int i = 0; i < top; i++) { edge[S[i]].flow += Min; edge[S[i]^1].flow -= Min; } ans += Min; top = inser; u = edge[S[top]^1].to; continue; } bool flag = false; int v; for(int i = cur[u]; i != -1; i = edge[i].next) { v = edge[i].to; if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]) { flag = true; cur[u] = i; break; } } if(flag) { S[top++] = cur[u]; u = v; continue; } int Min = N; for(int i = head[u]; i != -1; i = edge[i].next) if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min) { Min = dep[edge[i].to]; cur[u] = i; } gap[dep[u]]--; if(!gap[dep[u]])return ans; dep[u] = Min + 1; gap[dep[u]]++; if(u != start)u = edge[S[--top]^1].to; } return ans; } void init() { tol = 0; memset(head,-1,sizeof(head)); } //------------------------------------------------- struct Edge2 { int v; int cost; Edge2(int _v=0,int _cost=0):v(_v),cost(_cost){} }; vector<Edge2>E[MAXN],E2[MAXN]; void addedge(int u,int v,int w) { E[u].push_back(Edge2(v,w)); } bool vis[MAXN];//在队列标志 int cnt[MAXN];//每一个点的入队列次数 int dist[MAXN]; bool SPFA(int start,int n) { memset(vis,false,sizeof(vis)); for(int i=1;i<=n;i++) dist[i]=INF; vis[start]=true; dist[start]=0; queue<int>que; while(!que.empty())que.pop(); que.push(start); memset(cnt,0,sizeof(cnt)); cnt[start]=1; while(!que.empty()) { int u=que.front(); que.pop(); vis[u]=false; for(int i=0;i<E[u].size();i++){ int v=E[u][i].v; if(dist[v]>dist[u]+E[u][i].cost){ dist[v]=dist[u]+E[u][i].cost; if(!vis[v]){ vis[v]=true; que.push(v); if(++cnt[v]>n) return false; //cnt[i]为入队列次数,用来判定是否存在负环回路 } } } } return true; } set<int> s; int main() { int i,j,m,n; int a,b,c; while(scanf("%d %d",&n,&m)!=EOF) { init(); s.clear(); for(i=1;i<=n;i++) E[i].clear(); for(i=0; i<m; i++) { scanf("%d %d %d",&a,&b,&c); addedge(a,b,c); addedge(b,a,c); } int ok=SPFA(1,n); for(i=1;i<=n;i++){ E2[i]=E[i]; E[i].clear(); } for(i=1; i<=n; i++) { int sz=E2[i].size(); for(j=0; j<sz; j++) { int u=i; int v=E2[i][j].v; //得到最短路的边 if(dist[u]+E2[i][j].cost==dist[v]) { add(u,v,1); s.insert(u),s.insert(v); addedge(u,v,1); } } } int ans=sap(1,n,(int)s.size()); ok=SPFA(1,n); printf("%d %d\n",ans,m-dist[n]); } return 0; } /* 2 3 1 2 1 1 2 1 1 2 1 2 5 1 2 2 1 2 2 1 2 1 1 2 1 1 2 1 */
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css 来自http://css-tricks.com/snippets/css/media-queries-for-standard-devices/
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