ACM之路

【题目】http://poj.org/problem?id=1459 Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 23873   Accepted: 12456 Description A power network consi

Minimum Cost Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 15057   Accepted: 5176 Description Dearboy, a goods victualer, now comes to a big problem, and

Island Transport Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 6668    Accepted Submission(s): 2091 Problem Description 　　In th

Food Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4105    Accepted Submission(s): 1365 Problem Description 　　You, a part-time di

Max_Flow MF; int n,m; int main() { while(scanf("%d%d",&n,&m) != EOF) { if(n == 0 && m == 0) break; MF.Init(n+5,1,n); int u,v,w; for(int i=0;i<m;i++) { scanf("%d%d%d",&u,&v,&w); MF

int sumFlow; const int MAXN = 1010; const int MAXM = 1000200; const int INF = 1000000000; struct Edge { int u, v, cap, cost; int next; }edge[MAXM<<2]; int NE; int head[MAXN], dist[MAXN], pp[MA

上界改容量，下界先流走。  U -> V之间添加一条 cap - down的边 S->U 为U多出来的入的下界- 出的下界 V -> T 为V多出来的出的下界-入的下界 Max_Flow MF; int n,m; const int maxn = 210; const int maxm = 40010; int a[maxn]; int id[maxm]; int up[maxm]

Leapin' Lizards Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1752    Accepted Submission(s): 717 Problem Description Your platoo

Power Network Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 25417   Accepted: 13249 Description A power network consists of nodes (power stations, consum

Dining Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11698   Accepted: 5376 Description Cows are such finicky eaters. Each cow has a preference for certa

A Plug for UNIX Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14960   Accepted: 5062 Description You are in charge of setting up the press room for the i

#include #include #include #include using namespace std; const int maxn = 210; int N,M; int g[maxn][maxn]; int pre[maxn]; int cap[maxn]; int used[maxn]; const int INF = 0x3f3f3f3f; int EK() {

int n,m; Max_Flow MF; int main() { freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); while(scanf("%d%d",&n,&m) != EOF) { int S,T; scanf("%d%d",&S,&T); MF.Init(2*n+2,S,T