据说这道题目是个很经典的题,好多人测最大流算法效率都是用的这题,只会dinic的弱菜第一法果断tle了,把vector改成数组了时候5s过。
下次什么时候学了isap在写一遍把
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack> using namespace std; typedef long long LL; const int maxn = 20000 + 5; const int maxm = 200000 * 30; const LL INF = INT_MAX - 2; int first[maxn],nxt[maxm],v[maxm],cap[maxm]; int n,m,s,t,ans,ecnt; inline void adde(int uu,int vv,int ww) { v[ecnt] = vv; cap[ecnt] = ww; nxt[ecnt] = first[uu]; first[uu] = ecnt++; v[ecnt] = uu; cap[ecnt] = 0; nxt[ecnt] = first[vv]; first[vv] = ecnt++; } inline void input() { int a,b,c; for(int i = 1;i <= n;i++) { scanf("%d%d",&a,&b); adde(s,i,a); adde(i,t,b); } for(int i = 1;i <= m;i++) { scanf("%d%d%d",&a,&b,&c); adde(a,b,c); adde(b,a,c); } } int q[maxn * 2],qs,qe,level[maxn]; inline bool bfs() { qs = qe = 0; memset(level,0,sizeof level); q[qe++] = s; level[s] = 1; while(qs < qe) { int now = q[qs++]; if(now == t) break; for(int i = first[now];~i;i = nxt[i]) { if(!level[v[i]] && cap[i]) { level[v[i]] = level[now] + 1; q[qe++] = v[i]; } } } return level[t]; } int dfs(int now,int alpha) { if(now == t) return alpha; int sum = 0; for(int i = first[now];~i && alpha;i = nxt[i]) { if(level[v[i]] == level[now] + 1 && cap[i]) { int ret = dfs(v[i],min(alpha,cap[i])); cap[i] -= ret; cap[i ^ 1] += ret; sum += ret; alpha -= ret; } } if(sum == 0) level[now] = -1; return sum; } inline int dinic() { int ans = 0; while(bfs()) ans += dfs(s,INF); return ans; } inline void solve() { int ret = dinic(); printf("%d\n",ret); } int main() { while(scanf("%d%d",&n,&m) != EOF) { ecnt = 0; s = 0; t = n + 1; memset(first,-1,sizeof(first)); input(); solve(); } return 0; }
POJ 1637 Dual Core CPU 求最小割,布布扣,bubuko.com
原文:http://www.cnblogs.com/rolight/p/3871740.html