BZOJ 1706 usaco 2007 Nov relays 奶牛接力跑/POJ 3613 Cow Relays 倍增Floyd

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题目大意：求恰好走k步从S到T的最短路。

思路：设f[p][i][j]为从i到j恰好走2^p步的最短路，DP方程十分简单：f[p][i][j] = min(f[p][i][j],f[p - 1][i][k] + f[p - 1][k][j]);

对总步数T进行二进制拆分，在T有1的位置上，假如这个位置为p，那么就用f[p][][]来更新答案g[][],最后得到的g[][]就是答案矩阵。

注意要离散化一下。。

CODE：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 210using namespace std;#define min(a,b) ((a) < (b) ? (a):(b))#define max(a,b) ((a) > (b) ? (a):(b)) struct Complex{    int x,y,len;         void Read() {        scanf("%d%d%d",&len,&x,&y);    }}edge[MAX]; pair<int,int *> xx[1010];int cnt,t; int T,edges,points,start,end;int f[20][MAX][MAX],g[MAX][MAX],h[MAX][MAX]; int main(){    cin >> T >> edges >> start >> end;    memset(g,0x3f,sizeof(g));    memset(f,0x3f,sizeof(f));    for(int i = 1; i <= edges; ++i) {        edge[i].Read();        xx[++cnt].first = edge[i].x,xx[cnt].second = &edge[i].x;        xx[++cnt].first = edge[i].y,xx[cnt].second = &edge[i].y;    }    xx[++cnt].first = start,xx[cnt].second = &start;    xx[++cnt].first = end,xx[cnt].second = &end;    sort(xx + 1,xx + cnt + 1);    for(int i = 1; i <= cnt; ++i) {        if(i == 1 || xx[i].first != xx[i - 1].first)            ++t;        *xx[i].second = t;    }    for(int i = 1; i <= edges; ++i)        f[0][edge[i].x][edge[i].y] = f[0][edge[i].y][edge[i].x] = min(f[0][edge[i].x][edge[i].y],edge[i].len);    points = t;    for(int i = 1; i <= points; ++i)        g[i][i] = 0;    int p = 0;    while(T) {        if(T&1) {            memset(h,0x3f,sizeof(h));            for(int k = 1; k <= points; ++k)                for(int i = 1; i <= points; ++i)                    for(int j = 1; j <= points; ++j)                        h[i][j] = min(h[i][j],g[i][k] + f[p][k][j]);            memcpy(g,h,sizeof(g));        }        T >>= 1;        ++p;        for(int k = 1; k <= points; ++k)            for(int i = 1; i <= points; ++i)                for(int j = 1; j <= points; ++j)                    f[p][i][j] = min(f[p][i][j],f[p - 1][i][k] + f[p - 1][k][j]);    }    cout << g[start][end] << endl;    return 0;}

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