【模板】網路最大流
阿新 • • 發佈:2018-11-19
題目描述
如題,給出一個網路圖,以及其源點和匯點,求出其網路最大流。
輸入輸出格式
輸入格式:第一行包含四個正整數N、M、S、T,分別表示點的個數、有向邊的個數、源點序號、匯點序號。
接下來M行每行包含三個正整數ui、vi、wi,表示第i條有向邊從ui出發,到達vi,邊權為wi(即該邊最大流量為wi)
輸出格式:一行,包含一個正整數,即為該網路的最大流。
輸入輸出樣例
輸入樣例#1: 複製4 5 4 3 4 2 30 4 3 20 2 3 20 2 1 30 1 3 40輸出樣例#1: 複製
50
說明
時空限制:1000ms,128M
資料規模:
對於30%的資料:N<=10,M<=25
對於70%的資料:N<=200,M<=1000
對於100%的資料:N<=10000,M<=100000
樣例說明:
題目中存在3條路徑:
4-->2-->3,該路線可通過20的流量
4-->3,可通過20的流量
4-->2-->1-->3,可通過10的流量(邊4-->2之前已經耗費了20的流量)
故流量總計20+20+10=50。輸出50。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize("O3") using namespace std; #define maxn 200005 #define inf 0x3f3f3f3f #define INF 9999999999 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 1e9 + 7; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-3 typedef pair<int, int> pii; #define pi acos(-1.0) const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) typedef pair<int, int> pii; inline ll rd() { ll x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x; } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } ll sqr(ll x) { return x * x; } /*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans; } */ ll qpow(ll a, ll b, ll c) { ll ans = 1; a = a % c; while (b) { if (b % 2)ans = ans * a%c; b /= 2; a = a * a%c; } return ans; } int n, m; int st, ed; struct node { int u, v, nxt, w; }edge[maxn<<1]; int head[maxn], cnt; void addedge(int u, int v, int w) { edge[cnt].u = u; edge[cnt].v = v; edge[cnt].nxt = head[u]; edge[cnt].w = w; head[u] = cnt++; } int rk[maxn]; int bfs() { queue<int>q; ms(rk); rk[st] = 1; q.push(st); while (!q.empty()) { int tmp = q.front(); q.pop(); for (int i = head[tmp]; i != -1; i = edge[i].nxt) { int to = edge[i].v; if (rk[to] || edge[i].w <= 0)continue; rk[to] = rk[tmp] + 1; q.push(to); } } return rk[ed]; } int dfs(int u, int flow) { if (u == ed)return flow; int add = 0; for (int i = head[u]; i != -1 && add < flow; i = edge[i].nxt) { int v = edge[i].v; if (rk[v] != rk[u] + 1 || !edge[i].w)continue; int tmpadd = dfs(v, min(edge[i].w, flow - add)); if (!tmpadd) { rk[v] = -1; continue; } edge[i].w -= tmpadd; edge[i ^ 1].w += tmpadd; add += tmpadd; } return add; } int ans; void dinic() { while (bfs())ans += dfs(st, inf); } int main() { //ios::sync_with_stdio(0); memset(head, -1, sizeof(head)); rdint(n); rdint(m); rdint(st); rdint(ed); for (int i = 1; i <= m; i++) { int u, v, w; rdint(u); rdint(v); rdint(w); addedge(u, v, w); addedge(v, u, 0); } dinic(); cout << ans << endl; return 0; }