费用流 csu1506 Double Shortest Paths

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题意：Alice和Bob要从1走到n，每条边第一次被走过的时候的权值是d，第二次被走过的时候是d+a

思路：费用流裸题，对于将源点连到1，将n连到汇点，对于每条边建立两条边，一条费用为d,另一条费用为d+a然后跑一遍最小费用最大流答案就出来了

#include<map>#include<set>#include<cmath>#include<stack>#include<queue>#include<cstdio>#include<string>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define FIN freopen("input.txt","r",stdin)using namespace std;typedef long long LL;typedef pair<LL, int> PLI;const int MX = 2000 + 5;const int MM = 8000 + 5;const int INF = 0x3f3f3f3f;struct Edge {    int to, next, cap, flow, cost;    Edge() {}    Edge(int _to, int _next, int _cap, int _flow, int _cost) {        to = _to; next = _next; cap = _cap; flow = _flow; cost = _cost;    }} E[MM];int Head[MX], tol;int pre[MX]; //储存前驱顶点int dis[MX]; //储存到源点s的距离bool vis[MX];int N;//节点总个数，节点编号从0~N-1void init(int n) {    tol = 0;    N = 2 * n + 2;    memset(Head, -1, sizeof(Head));}void edge_add(int u, int v, int cap, int cost) {    E[tol] = Edge(v, Head[u], cap, 0, cost);    Head[u] = tol++;    E[tol] = Edge(u, Head[v], 0, 0, -cost);    Head[v] = tol++;}bool spfa(int s, int t) {    queue<int>q;    for (int i = 0; i < N; i++) {        dis[i] = INF;        vis[i] = false;        pre[i] = -1;    }    dis[s] = 0;    vis[s] = true;    q.push(s);    while (!q.empty()) {        int u = q.front();        q.pop();        vis[u] = false;        for (int i = Head[u]; i != -1; i = E[i].next) {            int v = E[i].to;            if (E[i].cap > E[i].flow && dis[v] > dis[u] + E[i].cost) {                dis[v] = dis[u] + E[i].cost;                pre[v] = i;                if (!vis[v]) {                    vis[v] = true;                    q.push(v);                }            }        }    }    if (pre[t] == -1) return false;    else return true;}//返回的是最大流， cost存的是最小费用int minCostMaxflow(int s, int t, int &cost) {    int flow = 0;    cost = 0;    while (spfa(s, t)) {        int Min = INF;        for (int i = pre[t]; i != -1; i = pre[E[i ^ 1].to]) {            if (Min > E[i].cap - E[i].flow)                Min = E[i].cap - E[i].flow;        }        for (int i = pre[t]; i != -1; i = pre[E[i ^ 1].to]) {            E[i].flow += Min;            E[i ^ 1].flow -= Min;            cost += E[i].cost * Min;        }        flow += Min;    }    return flow;}int main() {    int n, m, ansk = 0;//FIN;    while(~scanf("%d%d", &n, &m)) {        init(n);        int s = 0, t = 2 * n + 1;        edge_add(s, 1, 2, 0);        edge_add(n, t, 2, 0);        for(int i = 1; i <= m; i++) {            int u, v, d, a;            scanf("%d%d%d%d", &u, &v, &d, &a);            edge_add(u, v, 1, d);            edge_add(u, v, 1, d + a);        }        int ans = 0;        minCostMaxflow(s, t, ans);        printf("Case %d: %d\n", ++ansk, ans);    }    return 0;}

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