BZOJ 1001: [BeiJing2006] 狼抓兔子

来源：互联网发布：华为荣耀8数据编辑：程序博客网时间：2024/06/10 07:53
利用平面图的性质把最小割转化为最短路问题，大概是这样的：
太菜了。。。先是数组开太大，中间dijkstra的pair还写反了（捂脸）
/**************************************************************    Problem: 1001    User: will7101    Language: C++    Result: Accepted    Time:1912 ms    Memory:95340 kb****************************************************************/ #include <cstdio>#include <algorithm>#include <cstring>#include <utility>#include <queue>using namespace std; typedef long long ll; const int MAXV = 2200000, MAXE = 6600000, INF = 0x3f3f3f3f;struct edge{int to, cost, next; } es[MAXE];int head[MAXV], E, V, N, M, s, e; void init(){    memset(head, -1, sizeof(head));} inline void add2(const int &a, const int &b, const int &cost){    es[E] = (edge) {b, cost, head[a]};    head[a] = E++;    es[E] = (edge) {a, cost, head[b]};    head[b] = E++;} int d[MAXV];  int dijkstra(int s, int e){    typedef pair<int, int> P;    priority_queue<P, vector<P>, greater<P> > que;    memset(d, 0x3f, sizeof(d));    d[s] = 0;    que.push(P(0, s));    while(!que.empty()){        P p = que.top(); que.pop();        int u = p.second;        if(p.first > d[u]) continue;        for(int i = head[u]; i != -1; i = es[i].next){            int v = es[i].to;            if(d[v] > d[u] + es[i].cost){                d[v] = d[u] + es[i].cost;                que.push(P(d[v], v));            }        }    }    return d[e];} /* int SPFA(int s, int e){    queue<int> que;    memset(d, 0x3f, sizeof(d));    d[s] = 0;    que.push(s); vis[s] = true;         while(!que.empty()){        int u = que.front(); que.pop(); vis[u] = false;        for(int i = head[u]; i != -1; i = es[i].next){            int v = es[i].to, c = es[i].cost;            if(d[v] > d[u] + c){                 d[v] = d[u] + c;                if(!vis[v]){                    que.push(v);                    vis[v] = true;                }            }        }     }         return d[e];}*/ inline void rint(int &a){    char c;    while((c = getchar()) < '0' || c > '9');    a = c - '0';    while((c = getchar()) >= '0' && c <= '9') a = (a << 3) + (a << 1) + c - '0';} void read(){    int sz = (N - 1) * (M - 1), c;    s = sz * 2; e = s + 1;    for(int i = 0; i < M - 1; i++){        rint(c);        add2(s, i, c);    }    for(int i = 1; i < N - 1; i++){        for(int j = 0; j < M-1; j++){            rint(c);            add2(sz + (i - 1) * (M - 1) + j, i * (M - 1) + j, c);        }    }    for(int i = 0; i < M - 1; i++){        rint(c);        add2(sz + (N - 2) * (M - 1) + i, e, c);    }//==========================================================        for(int i = 0; i < N - 1; i++){        rint(c);        add2(e, sz + i * (M - 1), c);        for(int j = 0; j < M - 2; j++){            scanf("%d", &c);            add2(i * (M - 1) + j, sz + i * (M - 1) + j + 1, c);        }        rint(c);        add2(i * (M - 1) + M - 2, s, c);    }    for(int i = 0; i < N - 1; i++){        for(int j = 0; j < M - 1; j++){            rint(c);            add2(i * (M - 1) + j, sz + i * (M - 1) + j, c);        }    }} int main(){//  freopen("in.txt", "r", stdin);    scanf("%d%d", &N, &M);    if(N == 1 && M == 1){        printf("0\n");        return 0;    }    if(N == 1){        int ans = INF, c;        for(int i = 0; i < M - 1; i++){            scanf("%d", &c);            if(ans > c) ans = c;        }        printf("%d\n", ans);        return 0;    }    if(M == 1){        int ans = INF, c;        for(int i = 0; i < N - 1; i++){            scanf("%d", &c);            if(ans > c) ans = c;        }        printf("%d\n", ans);        return 0;    }    init();    read();    printf("%d\n", dijkstra(s, e));    return 0;}
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