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最新推荐文章于 2025-09-09 16:53:48 发布

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最新推荐文章于 2025-09-09 16:53:48 发布

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原文链接：http://www.cnblogs.com/ramanujan/p/3351176.html

There are N cities, and M directed roads connecting them. Now you want to transport K units of goods from city 1 to city N. There are many robbers on the road, so you must be very careful. The more goods you carry, the more dangerous it is. To be more specific, for each road i, there is a coefficient a_i. If you want to carry x units of goods along this road, you should pay a_i* x² dollars to hire guards to protect your goods. And what's worse, for each road i, there is an upper bound C_i, which means that you cannot transport more than C_i units of goods along this road. Please note you can only carry integral unit of goods along each road.

You should find out the minimum cost to transport all the goods safely.

Input

There are several test cases. The first line of each case contains three integers, N, M and K. (1 $\le$ N $\le$ 100, 1 $\le$ M $\le$ 5000, 0 $\le$ K $\le$ 100). Then M lines followed, each contains four integers (u_i, v_i, a_i, C_i), indicating there is a directed road from city u_i to v_i, whose coefficient is a_i and upper bound is C_i. (1 $\le$ u_i, v_i $\le$ N, 0 < a_i $\le$ 100, C_i $\le$ 5)

Output

Output one line for each test case, indicating the minimum cost. If it is impossible to transport all the K units of goods, output `-1'.

Sample Input

Sample Output

4 
-1 
3

书上讲过的处理方法，当时没看懂，做题时YY了出来。
因为cap<=5，所以有这几种：1， 4， 9， 16， 25
其实它们可分别表示为和的方式：（1）， （1+3）， （1+3+5），（1+3+5+7）， （1+3+5+7+9）
那么对以后的项是否也可这么做呢？
因为（i从1到n求和）（2*i-1） = (2*n-1 + 1) * n/2 = n^2

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<map>

using namespace std;

#define LL long long
#define ULL unsigned long long
#define UINT unsigned int
#define cint const int
#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define MAX(X,Y) ((X) > (Y) ? (X) : (Y))
#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))
#define INF 100000000
#define MAXN 110
#define MAXM 50010

struct edge{
    int u, v, cap, flow, cst, nxt;
}e[MAXM];
int h[MAXN], cc;

void add(int u, int v, int cap, int cst){
    e[cc]=(edge){u, v, cap, 0, cst, h[u]};
    h[u]=cc++;
    e[cc]=(edge){v, u, 0, 0, -cst, h[v]};
    h[v]=cc++;
}

int d[MAXN], p[MAXN], inq[MAXN];
int a[MAXN], k;
void mcmf(cint s, cint t, cint n, int &mincst, int &maxflow){
    mincst=maxflow=0;
    while(1){
        queue<int> q;   q.push(s);
        memset(inq, 0, sizeof(inq));
        for(int i=0; i<n; i++) d[i]=INF;
        inq[s]=1;   d[s]=0;
        a[s]=INF;
        while(!q.empty()){
            int u=q.front();    q.pop();    inq[u]=0;
            for(int i=h[u]; i!=-1; i=e[i].nxt){
                int v=e[i].v, cap=e[i].cap, ef=e[i].flow, cst=e[i].cst;
                if(d[v]>d[u]+cst && cap>ef){
                    d[v]=d[u]+cst;
                    a[v]=MIN(a[u], cap-ef);
                    p[v]=i;
                    if(!inq[v]) q.push(v), inq[v]=1;
                }
            }
        }
        if(d[t]==INF) break;
        maxflow+=a[t];
        mincst+=a[t]*d[t];
        for(int u=t; u!=s; u=e[p[u]].u){
            e[p[u]].flow+=a[t];
            e[p[u]^1].flow-=a[t];
        }
    }
}

int main(){
//    freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
    int n, m;
    while(scanf(" %d %d %d", &n, &m, &k)==3){
        int i, j, u, v, a, cap;
        memset(h, -1, sizeof(h));       cc=0;
        for(i=0; i<m; i++){
            scanf(" %d %d %d %d", &u, &v, &a, &cap);
            for(j=1; j<=cap; j++)
                add(u, v, 1, a*(2*j-1));
        }
        add(0, 1, k, 0);
        int mincst, maxflow;
        mcmf(0, n, n+1, mincst, maxflow);
        if(maxflow!=k) printf("-1\n");
        else printf("%d\n", mincst);
    }
    return 0;
}

转载于:https://www.cnblogs.com/ramanujan/p/3351176.html