Gym - 101908G Gasoline 二分+最大流

 

G - Gasoline Gym - 101908G

 

题意：给出R个提供点，P个接收点，每个接收点都要接收满，还有一个运输的时间，问最小时间能够完成所有的运输

题解：首先每次都必须要满流，所以我们只要限制时间即可，限制时间加边，跑最大流，如果最大流不是满流，那么就不满足，否则满足条件二分即可。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include<queue>
using namespace std;

#define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn =  1e5+5;
const int  mod = 1e9+7;

int P,R,C;
int sum;
int p[1005],r[1005];
struct E
{
    int u,v,w;
}e[20005];
struct Edge
{
    int to, cap, flow;
    int next;
}edge[4*maxn];
int tol;
int head[maxn];
int dep[maxn],pre[maxn],cur[maxn];
int gap[maxn];
void init()
{
    tol = 0;
    memset(head,-1,sizeof (head));
}

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;
    int i;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            for( 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( 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( 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( 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;
}

bool judge(int T) {
    init();
    for (int i = 1; i <= P; ++i) {
        addedge(2 * i - 1, 2 * i, p[i]);
        addedge(2 * i, 2 * P + 2 * R + 1, INF);
    }
    for (int i = 1; i <= C; ++i)
        if(e[i].w<=T)
            addedge(2*P+2*e[i].u,2*e[i].v-1,r[e[i].u]);
    for(int i=1;i<=R;++i)
    {
        addedge(0,2*P+2*i-1,INF);
        addedge(2*P+2*i-1,2*P+2*i,r[i]);
    }
    if(sap(0,2*P+2*R+1,2*P+2*R+2) == sum)
        return true;
    else
        return false;
}
int main()
{
    sum = 0;
    scanf("%d%d%d",&P,&R,&C);
    for(int i=1;i<=P;++i)
    {
        scanf("%d",p+i);
        sum += p[i];
    }
    for(int i=1;i<=R;++i)
        scanf("%d",r+i);
    for(int i=1;i<=C;++i)
        scanf("%d%d%d",&e[i].v,&e[i].u,&e[i].w);
    int l = 1,r = 1000005;
    bool flag = false;
    while(l < r)
    {
        int mid = (l+r)/2;
        if(judge(mid))
        {
            r = mid;
            flag = true;
        }
        else
            l  = mid + 1;
    }
    if(flag)
        printf("%d\n",l);
    else
        puts("-1");
}

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