bzoj 3396 [Usaco2009 Jan]Total flow 水流

Description

Input

第1行输入N，之后N行每行描述一条水管，前两个英文字母表示水管的两端（大小写字母是不一样的），后一个整数表示水管的流量，流量不会超过1000.

Output

一个整数，表示总流量．

Sample Input

5

A B 3

B C 3

C D 5

D Z 4

B Z 6
Sample Output

3
HINT

Source

Silver

用来练习网络流的模板题。
专属于权限的水题。

#include <bits/stdc++.h>
#define inf 0x7f7f7f7f
#define N 510
#define M 50010
using namespace std;
struct Edge
{
    int next, to, v;
}e[M];
int head[N], n, m, cnt = 1, cur[N], h[N], q[M], ans, T;

void ins(int u, int v, int w)
{
    e[++ cnt].to = v;
    e[cnt].next = head[u];
    e[cnt].v = w;
    head[u] = cnt;
}

inline int read()
{
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

void insert(int u, int v, int w)
{
    ins(u, v, w);
    ins(v, u, 0);
}

int bfs()
{
    int t = 0, w = 1;
    memset(h, -1, sizeof(h));
    h[1] = 0;
    q[0] = 1;
    while(t != w)
    {
        int now = q[t ++];
        if(t == M) t = 0;
        for(int i = head[now]; i; i = e[i].next)
        {
            int y = e[i].to;
            if(h[y] == -1 && e[i].v)
            {
                h[y] = h[now] + 1;
                q[w ++] = y;
                if(w == M) w = 0;
            }
        }
    }
    if(h[T] == -1) return 0;
    else return 1;
}

int dfs(int x, int f)
{
    if(x == T) return f;
    int used = 0, w;
    for(int i = cur[x]; i; i = e[i].next)
    {
        int y = e[i].to;
        if(h[y] == h[x] + 1 && e[i].v)
        {
            w = dfs(y, min(e[i].v, f - used));
            used += w;
            e[i].v -= w;
            if(e[i].v) cur[x] = i;
            e[i ^ 1].v += w;
            if(used == f) return f;
        }
    }
    if(!used) h[x] = -1;
    return used;
}


void dinic()
{
    while(bfs())
    {
        for(int i = 0; i <= 60; i ++)
            cur[i] = head[i];
        ans += dfs(1, inf);
    }
}

int main()
{
    n = read();
    T = 26;
    for(int i = 1; i <= n; i ++)
    {
        char ch[5];
        int a,b;
        scanf("%s",ch);
        if(ch[0] >= 'A' && ch[0] <= 'Z') a = ch[0] - 'A' + 1;
        else a = ch[0] - 'a' + 27;
        scanf("%s",ch);
        if(ch[0] >= 'A' && ch[0] <= 'Z') b = ch[0] - 'A' + 1;
        else b = ch[0] - 'a' + 27;
        int x = read();
        insert(a,b,x);
    }
    dinic();
    printf("%d",ans);
    return 0;
}