題意:
(from luckycat)
解法: 題目已經非常明顯是求 s-t 最大流囉。
TAG: Flow Network, Ford-Fulkerson, MaxFlow
注意
程式碼:
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#include <iostream> | |
#include <cstdio> | |
#include <cstring> | |
using namespace std; | |
#define clr( x, v ) memset( x, v, sizeof(x) ) | |
#define N 105 | |
#define INF 1e7 | |
int n, casei = 1; | |
int s, t, c; | |
// maximum flow algorithm | |
int F[N][N]; | |
int C[N][N]; | |
int pre[N], que[N], d[N], mk[N]; | |
int maxflow( void ) | |
{ | |
static int head, tail; | |
int i, u, v, flow; | |
flow = 0; | |
clr( F, 0 ); | |
do { | |
clr( mk, 0 ); clr( d, 0 ); | |
que[0] = s; mk[s] = 1; d[s] = INF; | |
for( pre[s]=head=tail=0; head<=tail && !mk[t]; ++head ) { | |
for( u=que[head], v=1; v <= n; ++v ) if( !mk[v] && F[u][v]<C[u][v] ) { | |
mk[v] = 1; que[++tail] = v; pre[v] = u; | |
if( d[u] < C[u][v]-F[u][v] ) d[v] = d[u]; | |
else d[v] = C[u][v]-F[u][v]; | |
} | |
} | |
if( !mk[t] ) break; flow += d[t]; | |
for( u=t; u!=s; ) { v = u; u = pre[v]; F[u][v] += d[t]; F[v][u] = -F[u][v]; } | |
} while( d[t] > 0 ); return flow; | |
} | |
int main( void ) | |
{ | |
int i, a, b, v; | |
while( scanf( "%d", &n ) == 1 && n ) { | |
printf( "Network %d\n", casei++ ); | |
// input & initialization | |
scanf( "%d%d%d", &s, &t, &c ); | |
clr( C, 0 ); | |
for( i = 1; i <= c; ++i ) { | |
scanf( "%d%d%d", &a, &b, &v ); | |
C[a][b] = C[b][a] += v; | |
} | |
// solve & output | |
printf( "The bandwidth is %d.\n\n", maxflow() ); | |
} | |
return 0; | |
} |
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