poj 2391 二分+多源最短路+最大流

poj 2391

题意：有n个挡雨棚，每个挡雨棚可以容纳Bi头牛，现在每个挡雨棚下有Ai头牛，有m条路，每条路连接两个挡雨棚Ui和Vi，每头牛经过这条路需要花费Ti的时间，忽略牛的碰撞体积～

思路：

这种求时间和流量没什么关系的题，可以通过二分枚举时间往图里增边，判断是否满流来求时间，这就需要用floyd求多源最短路。

建图：显然挡雨棚的容量是流量来源，牛的数量是最大流量，所以每个挡雨棚都要和源点汇点建边，挡雨棚和挡雨棚之间也要拆点，当最短路径小于等于二分限制长度时建边，边的容量为INF。

拆点原因：如果不拆会出现点间接连通的问题，说的形象一点，对于点A、B、C，A-B为a，B-C为b（b>=a），显然A-C为a+b，显然如果将长度限制在b，那么A-C是没有的，但是A可以通过A-B，B-C到达C，这是我们所不希望的。

另外二分的时候有个细节搞了好久。。。二分写挫了吧右边界是无法到达的，这地方卡了好久。。。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MAXN = 410;//点数的最大值
const int MAXM = 100100;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge{
int to,next, cap, flow;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init(){
tol = 0;
memset(head,-1,sizeof(head));
}
//加边，单向图三个参数，双向图四个参数
void addedge(int u,int v,int w,int rw=0){
edge[tol].to = v;edge[tol].cap = w;edge[tol].next = head[u];
edge[tol].flow = 0;head[u] = tol++;
edge[tol].to = u;edge[tol].cap = rw;edge[tol].next = head[v];
edge[tol].flow = 0;head[v]=tol++;
}
int sap(int start,int end,int N){
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,head,sizeof(head));
int u = start;
pre[u] = -1;
gap[0] = N;
int ans = 0;
while(dep[start] < N){
if(u == end){
int Min = INF;
for(int i = pre[u];i != -1; i = pre[edge[i^1].to])
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
for(int i = pre[u];i != -1; i = pre[edge[i^1].to]){
edge[i].flow += Min;
edge[i^1].flow -= Min;
}
u = start;
ans += Min;
continue;
}
bool flag = false;
int v;
for(int i = cur[u]; i != -1;i = edge[i].next){
v = edge[i].to;
if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]){
flag = true;
cur[u] = pre[v] = i;
break;
}
}
if(flag){
u = v;
continue;
}
int Min = N;
for(int i = head[u]; i != -1;i = edge[i].next)
if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min){
Min = dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if(!gap[dep[u]])return ans;
dep[u] = Min+1;
gap[dep[u]]++;
if(u != start) u = edge[pre[u]^1].to;
}
return ans;
}
int cap[MAXN], now[MAXN];
long long dist[MAXN][MAXN];
main() {
int n, m;
while(~scanf("%d %d", &n, &m)){
memset(dist, 0, sizeof dist);
for(int i = 0; i < MAXN; i++)
for(int j = 0; j < MAXN; j++)
dist[i][j] = i == j? 0 : 0x7fffffffffffff;
int sum = 0;
long long maxtime = 0;
for(int i = 1; i <= n; i++) scanf("%d %d", &now[i], &cap[i]), sum += now[i];
for(int i = 1; i <= m; i++) {
int a, b;
long long c;
scanf("%d %d %I64d", &a, &b, &c);
dist[a][b] = dist[b][a] = min(dist[b][a], c);
maxtime += c;
}
for(int k = 1; k <= n; k++)
for(int i = 1; i <= n; i++)
if(dist[i][k] < 0x7fffffffffffff)
for(int j = 1; j <= n; j++)
dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
long long l = 1, r = maxtime + 10, mid, ans = -1;
int src = 0, sink = n + n + 1, res;
while(l < r){
mid = (l + r) / 2;
init();
for(int i = 1; i <= n; i++) {
addedge(src, i, now[i]);
addedge(n + i, sink, cap[i]);
addedge(i, i + n, INF);
}
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
if(dist[i][j] <= mid)
addedge(i, n + j, INF);
res = sap(src, sink, n + n + 2);
if(res == sum) r = ans = mid;
else l = mid + 1;
}
printf("%I64d\n", ans);
}
}