POJ-2391 Ombrophobic Bovines (二分答案+Floyd+拆点+最大流)

题目链接：https://cn.vjudge.net/contest/176665#problem/G

对网络流进行二分答案，总觉得容易超时，，，没勇气。

题解：

先通过一遍Floyd，得到任意两点间的最短距离，不走冤枉路，汇点T，每一个filed有一个have->已有的牛的数量，can->能放得下的牛，因为时间相互覆盖，所以，直接计算时间比较有困难性，发现有二分性质，短的时间可以，那么更长的时间一定可以，所以进行二分答案。

二分答案为时间，然后验证答案的正确性，建立网络流，S指向个点，容量为have，各点指向T，容量为can，也就表示出了有sum（have）的牛，跑向了各个点，然后又分配到了各个点，因为有时间答案的限制，所以，只能添加两点最短距离小于答案的边到图中，这样，完整的图就建立出来，然后跑最小割，如果最小割等于sum(have)，也就是满流，那么这个答案可行，缩小答案范围，继续验证。

注意，需要拆点，否则会发生串流，2->3 and 3->4 != 2->4。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long LL;
const LL N = 510;
const LL M = 540010;
const LL INF = 0x3f3f3f3f3f3f3f3f;

struct edge {
LL to, next;
LL c, f;
}graph[M];
LL totlen;
LL head
;

LL n, m;
LL have
, can
;
LL e

;
LL s, t;
LL sum;

void init(){
totlen = 0;
memset(head, -1, sizeof head);
}

void addEdge(LL u, LL v, LL w) {
graph[totlen] = {v, head[u], w, 0};
head[u] = totlen++;
graph[totlen] = {u, head[v], 0, 0};
head[v] = totlen++;
}

void Floyd(LL n) {
for(LL k = 1; k <= n; k++) {
for(LL i = 1; i <= n; i++) {
for(LL j = 1; j <= n; j++) {
e[i][j] = min(e[i][j], e[i][k]+e[k][j]);
}
}
}
}

// Dinic
LL que
;
LL cur
;
LL level
;
bool dinic_bfs(LL s, LL t) {
LL front = 0, tail = 0;
memset(level, 0, sizeof level);
level[s] = 1;
que[tail++] = s;
while(front != tail) {
LL u = que[front++];
if(front == N) front = 0;
for(LL i = head[u]; i != -1; i = graph[i].next) {
LL v = graph[i].to;
if(!level[v] && graph[i].c-graph[i].f > 0) {
level[v] = level[u]+1;
que[tail++] = v;
if(tail == N) tail = 0;
}
}
}
return level[t];
}

// cpflow: can pass flow  到达u点最大能通过的流量
LL dinic_dfs(LL u, LL t, LL cpflow) {
if(u == t) return cpflow;  // 到达汇点
// u 点到其他点 最多能增广的流量， 最多不能超过cpflow，由前面的边限制
LL addflow = 0;
for(LL& i = cur[u]; i != -1; i = graph[i].next) {
LL v = graph[i].to;
if(level[v] == level[u]+1 && graph[i].c-graph[i].f > 0) {
addflow = dinic_dfs(v, t, min(graph[i].c-graph[i].f, cpflow));
if(addflow > 0) {
graph[i].f += addflow;  // 正向通过的流量加
graph[i^1].f -= addflow;  // 反向的流量就得减
return addflow;        // 增广成功，直接返回。
}
}
}
return addflow;
}

LL dinic(LL s, LL t) {
LL maxflow = 0;
LL addflow = 0;
while(dinic_bfs(s, t)) {
memcpy(cur, head, sizeof head);  // 提高效率
while((addflow = dinic_dfs(s, t, INF)) > 0)
maxflow += addflow;
}
return maxflow;
}

bool check(LL mid) {
init();
for(LL i = 1; i <= n; i++) {
addEdge(s, i, have[i]);
addEdge(i+n, t, can[i]);
addEdge(i, i+n, sum);
}
for(LL i = 1; i <= n; i++) {
for(LL j = i+1; j <= n; j++) {
if(e[i][j] <= mid) {
addEdge(i, j+n, sum);
addEdge(j, i+n, sum);
}
}
}
LL ans = dinic(s, t);
return ans >= sum;
}

int main() {
// cout << 0x3f3f3f3f << endl;
// cout << 0x3f3f3f3f3f3f3f3f << endl;
while(scanf("%lld%lld", &n, &m) != EOF) {
sum = 0;
s = 0, t = n*2+1;
for(int i = 0; i <= n; i++) {
for(int j = 0; j <= n;j ++) {
e[i][j] = INF;
}
}
// cout << e[0][0] << " " << e[0][1] << endl;

for(LL i = 1; i <= n; i++) {
scanf("%lld%lld", &have[i], &can[i]);
sum += have[i];
}
for(LL i = 1; i <= m; i++) {
LL a, b, c;
scanf("%lld%lld%lld", &a, &b, &c);
if(c < e[a][b])
e[a][b] = e[b][a] = c;
}

Floyd(n);
LL l = 1, r = INF-1;
LL ans = -1;
while(l <= r) {
LL mid = (l+r)/2;
if(check(mid)) {
r = mid-1;
ans = mid;
}else{
l = mid+1;
}
// cout << l << " " << r << " " << mid << endl;
}
printf("%lld\n", ans);
}
return 0;
}