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POJ 1637 Dual Core CPU 求最小割
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;}
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