poj3469 Dinic

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Dinic 模板

以后做题就用这个模板了。

题意：一台双核电脑，给你多个任务，分别给出每个任务在第一个核和第二个核上运行的消耗。后面的m行输入是给出两个任务在两个不同核上运行需要付出的额外消耗。

建图：把每个任务作为节点，在超级源点与任务间的连一条边，其容量为给任务在核1上运行的消耗，在该任务节点与超级汇点之间连一条边，容量为该任务在核2上运行的消耗。

在任务之间连接无向边，容量为两个任务不在同一核上运行的额外消耗。

//6132K 2875ms#include <iostream>using namespace std;const int nMax = 20030;const int mMax = 230000;const int inf = 0x7fffffff;int head[nMax], temp[nMax];int que[nMax], level[nMax];int ne;struct Edge{int v, c, next;} edges[2*(nMax+mMax)];void addEdges(int from, int to, int val1, int val2){   edges[ne].v = to;   edges[ne].c = val1;   edges[ne].next = head[from];   head[from] = ne++;   edges[ne].v = from;   edges[ne].c = val2;   edges[ne].next = head[to];   head[to] = ne++;}int Dinic(int sta, int end){int max = 0;    int i, k, curr;    while(true){memset(level, -1, sizeof(level));    int f, r;        f = 0; r = 1;que[0] = sta;level[sta] = 0;        while(f < r){curr = que[f++];for(i = head[curr]; i; i = edges[i].next)if(edges[i].c && level[k = edges[i].v] == -1){level[k] = level[curr] + 1;que[r++] = k;if(k == end){ f = r; break; }}}if(level[end] == -1) break;        memcpy(temp, head, sizeof(temp));int top;for(curr = sta, top = 0; ;){if(curr == end){  int tp = inf;   for(i = 0; i < top; i++)                    if(tp > edges[que[i]].c)tp = edges[que[f = i]].c;for(i = 0; i < top; i++){edges[que[i]].c -= tp;edges[que[i]^1].c += tp;}top = f;                max += tp; curr = edges[que[top-1]].v;}for(i = temp[curr]; temp[curr]; i = temp[curr] = edges[temp[curr]].next)if(edges[i].c && level[edges[i].v] == level[curr] + 1)break;            if(temp[curr]){que[top++] = temp[curr];    curr = edges[temp[curr]].v;}else{if(top == 0) break;level[curr] = -1;if((top--) == 1)curr = sta;else    curr = edges[head[top-1]].v;}}}return max;}int main(){int n, m;    int k, j, val1, val2;    int i;    //freopen("a.txt", "r", stdin);ne = 2;memset(head, 0, sizeof(head));scanf("%d%d", &n, &m);for(i = 1; i <= n; i++){       scanf("%d%d", &val1, &val2);   addEdges(0, i, val1, 0);   addEdges(i, n+1, val2, 0);}for(i = 1; i <= m; i++){scanf("%d%d%d", &k, &j, &val1);        addEdges(k, j, val1, val1);}printf("%d\n", Dinic(0, n+1));return 0;}