BZOJ 1570: [JSOI2008]Blue Mary的旅行( 二分答案 + 最大流 )

二分答案, 然后对于答案m, 把地点分成m层, 对于边(u, v), 第x层的u -> 第x+1层的v 连边. 然后第x层的u -> 第x+1层的u连边(+oo), S->第一层的1(PEOPLE_NUMBER), 每一层N -> T(+oo), 假如最大流是等于人数,就是可行答案. 

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#include<cstdio>
#include<cstring>
#include<algorithm>
 
using namespace std;
 
#define Id(a, b) ((a) + (b) * N)
 
const int maxn = 59;
const int maxV = 5009;
 
int V, S, T, cnt[maxV], h[maxV];
int d[maxn][maxn], NUM, N;
 
struct edge {
int t, c;
edge *n, *r;
} E[100000], *pt, *e, *H[maxV], *p[maxV], *cur[maxV];
 
inline void Add(int u, int v, int c) {
pt->t = v, pt->c = c, pt->n = H[u], H[u] = pt++;
}
inline void AddEdge(int u, int v, int c) {
Add(u, v, c), Add(v, u, 0);
H[u]->r = H[v], H[v]->r = H[u];
}
 
int maxFlow() {
for(int i = 0; i < V; i++)
cur[i] = H[i], cnt[i] = h[i] = 0;
cnt[0] = V;
int ret = 0;
for(int x = S, A = maxV; h[S] < V; ) {
for(e = cur[x]; e; e = e->n)
if(e->c && h[e->t] + 1 == h[x]) break;
if(e) {
A = min(e->c, A);
p[e->t] = cur[x] = e;
if((x = e->t) == T) {
for(; x != S; x = p[x]->r->t)
p[x]->c -= A, p[x]->r->c += A;
ret += A;
A = maxV;
}
} else {
if(!--cnt[h[x]]) break;
h[x] = V;
for(e = H[x]; e; e = e->n) if(h[e->t] + 1 < h[x] && e->c) {
h[x] = h[e->t] + 1;
cur[x] = e;
}
++cnt[h[x]];
if(x != S) x = p[x]->r->t;
}
}
return ret;
}
 
void Build(int x) {
pt = E;
memset(H, 0, sizeof H);
V = N * x, S = V++, T = V++;
AddEdge(S, 0, NUM);
for(int i = 0; i < x; i++)
AddEdge(Id(N - 1, i), T, maxV);
for(int i = 1; i < x; i++)
for(int j = 0; j < N; j++) {
for(int k = 0; k < N; k++)
if(d[j][k]) AddEdge(Id(j, i - 1), Id(k, i), d[j][k]);
AddEdge(Id(j, i - 1), Id(j, i), maxV);
}
}
 
void Init() {
int m;
scanf("%d%d%d", &N, &m, &NUM);
memset(d, 0, sizeof d);
while(m--) {
int u, v;
scanf("%d%d", &u, &v);
scanf("%d", &d[--u][--v]);
}
}
 
void Work() {
int l = 1, r = 100, ans;
while(l <= r) {
int m = (l + r) >> 1;
Build(m);
if(maxFlow() == NUM) {
ans = m, r = m - 1;
} else
l = m + 1;
}
printf("%d\n", --ans);
}
 
int main() {
Init();
Work();
return 0;
}

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posted @ 2016-02-03 12:40  JSZX11556  阅读(116)  评论(0编辑  收藏