HDU-3035-War

来源：互联网发布：算法培训班编辑：程序博客网时间：2024/05/19 15:42

ACM模版

描述

题解

这是我做过建图最复杂的题，单看问题，感觉是最小割（最大流），但是点数数百万，根本搞不定啊，所以可以转化成最短路来做，提供一个比较好的相关知识的PPT——浅析最大最小定理在信息学竞赛中的应用……我也刚接触这个，明天进一步学习，要睡了，医生说，以后不能再熬夜了，不然……~~~/(ㄒoㄒ)/~~

代码

#include <cstdio>#include <queue>#include <cstring>#include <iostream>using namespace std;const int MAXN = 1000100;const int MAXM = 7000000;const int INF = 0x3f3f3f3f;struct node{    int v, d;    node(int _v, int _d) : v(_v), d(_d) {}    bool operator < (const node a) const    {        return a.d < d;    }};int n, m, en, S, T;int head[MAXN];struct Edge{    int v, w, next;    Edge() {}    Edge(int _v, int _w, int _next) : v(_v), w(_w), next(_next) {}} e[MAXM];inline void add(int u, int v, int w){    e[en].v = v;    e[en].w = w;    e[en].next = head[u];    head[u] = en++;    e[en].v = u;    e[en].w = w;    e[en].next = head[v];    head[v] = en++;}void initGraph(){    en = S = 0;    T = n * m * 4 + 1;    for (int i = S; i <= T; i++)    {        head[i] = -1;    }    int x, y, z;    for (int j = 0; j < m; j++)    {        x = j * 4 + 1;        scanf("%d", &z);        add(S, x, z);    }    for (int i = 1; i < n; i++)    {        for (int j = 0; j < m; j++)        {            y = (i * m + j) * 4 + 1;            x = y - m * 4 + 3;            scanf("%d", &z);            add(x, y, z);        }    }    for (int j = 0; j < m; j++)    {        x = ((n - 1) * m + j) * 4 + 4;        scanf("%d", &z);        add(x, T, z);    }    for (int i = 0; i < n; i++)    {        scanf("%d", &z);        add(i * m * 4 + 2, T, z);        for (int j = 1; j < m; j++)        {            y = (i * m + j) * 4 + 2;            x = y - 3;            scanf("%d", &z);            add(x, y, z);        }        scanf("%d", &z);        add(S, (i + 1) * m * 4 - 1, z);    }    for (int i = 0; i < n; i++)    {        for (int j = 0; j < m; j++)        {            x = (i * m + j) * 4 + 1;            scanf("%d", &z);            add(x, x + 1, z);            scanf("%d", &z);            add(x, x + 2, z);        }        for (int j = 0; j < m; j++)        {            x = (i * m + j) * 4 + 4;            scanf("%d", &z);            add(x, x - 2, z);            scanf("%d", &z);            add(x, x - 1, z);        }    }}int dis[MAXN];bool vis[MAXN];void dijstra(){    for (int i = S; i <= T; i++)    {        dis[i] = INF;        vis[i] = 0;    }    dis[S] = 0;    priority_queue<node> q;    q.push(node(S,0));    while (!q.empty())    {        int u = q.top().v;        q.pop();        if (u == T)        {            break;        }        if (vis[u])        {            continue;        }        vis[u] = true;        for (int i = head[u]; i != -1; i = e[i].next)        {            int v = e[i].v;            if (!vis[v] && dis[v] > dis[u] + e[i].w)            {                dis[v] = dis[u] + e[i].w;                q.push(node(v, dis[v]));            }        }    }    printf("%d\n", dis[T]);}int main(){    while (~scanf("%d%d", &n, &m))    {        initGraph();        dijstra();    }    return 0;}

0 0