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 50010using 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;}