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MAXMint 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);    }}